Remarks concerning a method for accelerating the convergence of sequences

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A.C. Muresan
Institutul de Calcul

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A.C. Mureṣan, Remarks concerning a method for accelerating the convergence of sequences, Rev. Anal. Numer. Theor. Approx., 23 (1994) no. 1, pp. 79-87.

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Revue d’Analyse Numérique et de Théorie de l’Approximation

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Editura Academiei Române

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https://ictp.acad.ro/jnaat/journal/article/view/1999-vol28-no2-art3

 

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[1] Brezinski, C., A new approach to convergence acceleration methods. In: Cuyt, A. (ed.). Nonlinear numerical methods and rational approximation, pp. 373-405, Dordrecht: Reidel (1987), https://doi.org/10.1007/978-94-009-2901-2_21
[2] Matos, A.C., Acceleration methods based on convergence tests, Numer. Math., 58 (1990), pp. 329-340, https://doi.org/10.1007/bf01385628
[3] Ney A., A general asymptotic formula for evaluation of the rest of the convergenc series with positive terms (in Romanian), Studii şi cercetări de matematică, 12, 2, (1961), pp. 315-332.
[4] Ney, A., Observations concerning a convergence theorem for series with positive terms (in Romanian). Studii şi cercetări matematice, 22, 5 (1970), pp. 763-767.
[5] Rosculet, M.N., A convergence criterion for series with positive terms (in Romanian), Studii şi cercetări matematice, 19, 1 (1967), pp. 9-11.

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REMARKS CONCERNING A METHOD FOR ACCELERATING THE CONVERGENCE OF SEQUENCES

ADRIAN MURESAN(Cluj-Napoca)

1. INTRODUCTION

Let ( S n ) S n (S_(n))\left(S_{n}\right)(Sn) be a sequence of numbers converging to S S SSS. A sequence transformation consists in transforming the sequence ( S n S n S_(n)S_{n}Sn ) into the sequence ( T n T n T_(n)T_{n}Tn ), where
T n = T ( S n , S n + 1 , … , S n + p ) , n = 0 , 1 , … T n = T S n , S n + 1 , … , S n + p , n = 0 , 1 , … T_(n)=T(S_(n),S_(n+1),dots,S_(n+p)),n=0,1,dotsT_{n}=T\left(S_{n}, S_{n+1}, \ldots, S_{n+p}\right), n=0,1, \ldotsTn=T(Sn,Sn+1,…,Sn+p),n=0,1,…
and p p ppp is a fixed integer. The aim of such transformation is to provide a sequence ( T n T n T_(n)T_{n}Tn ) converging to S S SSS faster than ( S n S n S_(n)S_{n}Sn ), that is,
lim n → ∞ T n − S S n − S = 0 . lim n → ∞   T n − S S n − S = 0 . lim_(n rarr oo)(T_(n)-S)/(S_(n)-S)=0.\lim _{n \rightarrow \infty} \frac{T_{n}-S}{S_{n}-S}=0 .limn→∞Tn−SSn−S=0.
It is known (see [1]) that accelerating convergence is equivalent to finding a "perfect estimation of the error", that is, a sequence ( D n D n D_(n)D_{n}Dn ) such that:
lim n → ∞ D n S − S n = 1 lim n → ∞   D n S − S n = 1 lim_(n rarr oo)(D_(n))/(S-S_(n))=1\lim _{n \rightarrow \infty} \frac{D_{n}}{S-S_{n}}=1limn→∞DnS−Sn=1
Proceeding in this way, Ana C. Matos proposed in [2] an accelerating method based on a general convergence test and on the following reasoning :
Let ( S n ) S n (S_(n))\left(S_{n}\right)(Sn) be a monotone sequence and ( x n ) x n (x_(n))\left(x_{n}\right)(xn) a given auxiliary sequence converging to a known finite limit x x xxx.
Let us define:
(1) r n := x − x n , ∀ n ∈ N (1) r n := x − x n , ∀ n ∈ N {:(1)r_(n):=x-x_(n)","AA n inN:}\begin{equation*} r_{n}:=x-x_{n}, \forall n \in \mathbb{N} \tag{1} \end{equation*}(1)rn:=x−xn,∀n∈N
(2) A n ( k ) := Δ x n Δ S n + k , ∀ n ∈ N , k ∈ N (2) A n ( k ) := Δ x n Δ S n + k , ∀ n ∈ N , k ∈ N {:(2)A_(n)(k):=(Deltax_(n))/(DeltaS_(n+k))","quad AA n inN","quad k inN:}\begin{equation*} A_{n}(k):=\frac{\Delta x_{n}}{\Delta S_{n+k}}, \quad \forall n \in \mathbb{N}, \quad k \in \mathbb{N} \tag{2} \end{equation*}(2)An(k):=ΔxnΔSn+k,∀n∈N,k∈N
where Δ x n = x n + 1 − x n , Δ S n + k = S n + k + 1 − S n + k Δ x n = x n + 1 − x n , Δ S n + k = S n + k + 1 − S n + k Deltax_(n)=x_(n+1)-x_(n),DeltaS_(n+k)=S_(n+k+1)-S_(n+k)\Delta x_{n}=x_{n+1}-x_{n}, \Delta S_{n+k}=S_{n+k+1}-S_{n+k}Δxn=xn+1−xn,ΔSn+k=Sn+k+1−Sn+k, and consider the following convergence test for sequences :
If,
(3) lim n → ∞ inf A n ( k ) > 0 (3) lim n → ∞   inf A n ( k ) > 0 {:(3)lim_(n rarr oo)i n fA_(n)(k) > 0:}\begin{equation*} \lim _{n \rightarrow \infty} \inf A_{n}(k)>0 \tag{3} \end{equation*}(3)limn→∞infAn(k)>0
or
(4) lim n → ∞ sup A n ( k ) < 0 (4) lim n → ∞   sup A n ( k ) < 0 {:(4)lim_(n rarr oo)s u pA_(n)(k) < 0:}\begin{equation*} \lim _{n \rightarrow \infty} \sup A_{n}(k)<0 \tag{4} \end{equation*}(4)limn→∞supAn(k)<0
then ( S n S n S_(n)S_{n}Sn ) converges.
This test enables us to obtain an estimation for ( S − S n ) S − S n (S-S_(n))\left(S-S_{n}\right)(S−Sn). In fact, if we suppose that ∀ n ∈ N , Δ S n > 0 ∀ n ∈ N , Δ S n > 0 AA n inN,DeltaS_(n) > 0\forall n \in \mathbb{N}, \Delta S_{n}>0∀n∈N,ΔSn>0 (the case when Δ S n < 0 , ∀ n ∈ N Δ S n < 0 , ∀ n ∈ N DeltaS_(n) < 0,AA n inN\Delta S_{n}<0, \forall n \in \mathbb{N}ΔSn<0,∀n∈N can be treated in the same way and if conditions (3) are satisfied, then we have:
∃ N ∈ N , ∀ n ∈ N , sup j → n A j ( k ) ⩾ Δ x m Δ S m + k ⩾ inf j → n A j ( k ) > 0 , ∀ m ⩾ n ∃ N ∈ N , ∀ n ∈ N , sup j → n   A j ( k ) ⩾ Δ x m Δ S m + k ⩾ inf j → n   A j ( k ) > 0 , ∀ m ⩾ n EE N inN,quad AA n inN,quads u p_(j rarr n)A_(j)(k) >= (Deltax_(m))/(DeltaS_(m+k)) >= i n f_(j rarr n)A_(j)(k) > 0,quad AA m >= n\exists N \in \mathbb{N}, \quad \forall n \in \mathbb{N}, \quad \sup _{j \rightarrow n} A_{j}(k) \geqslant \frac{\Delta x_{m}}{\Delta S_{m+k}} \geqslant \inf _{j \rightarrow n} A_{j}(k)>0, \quad \forall m \geqslant n∃N∈N,∀n∈N,supj→nAj(k)⩾ΔxmΔSm+k⩾infj→nAj(k)>0,∀m⩾n,
and since Δ S n > 0 , ∀ n ∈ N Δ S n > 0 , ∀ n ∈ N DeltaS_(n) > 0,AA n in N\Delta S_{n}>0, \forall n \in NΔSn>0,∀n∈N, we get Δ x m > 0 , ∀ m ⩾ n , ∀ n ⩾ N Δ x m > 0 , ∀ m ⩾ n , ∀ n ⩾ N Deltax_(m) > 0,AA m >= n,AA n >= N\Delta x_{m}>0, \forall m \geqslant n, \forall n \geqslant NΔxm>0,∀m⩾n,∀n⩾N, hence
Δ x m sup j ⩾ n A j ( k ) ⩽ Δ S m + k ⩽ Δ x m inf j ⩾ n A j ( k ) , ∀ n ⩾ N , ∀ m ⩾ n Δ x m sup j ⩾ n   A j ( k ) ⩽ Δ S m + k ⩽ Δ x m inf j ⩾ n   A j ( k ) , ∀ n ⩾ N , ∀ m ⩾ n (Deltax_(m))/(s u p_(j >= n)A_(j)(k)) <= DeltaS_(m+k) <= (Deltax_(m))/(i n f_(j >= n)A_(j)(k)),AA n >= N,AA m >= n\frac{\Delta x_{m}}{\sup _{j \geqslant n} A_{j}(k)} \leqslant \Delta S_{m+k} \leqslant \frac{\Delta x_{m}}{\inf _{j \geqslant n} A_{j}(k)}, \forall n \geqslant N, \forall m \geqslant nΔxmsupj⩾nAj(k)⩽ΔSm+k⩽Δxminfj⩾nAj(k),∀n⩾N,∀m⩾n
By adding these inequalities, member by member, for m = n , n + 1 , … m = n , n + 1 , … m=n,n+1,dotsm=n, n+1, \ldotsm=n,n+1,… we get
(5) r n sup j → n A j ( k ) ⩽ R n + k ⩽ r n inf j ⩾ n A j ( k ) , ∀ n ⩾ N , (5) r n sup j → n   A j ( k ) ⩽ R n + k ⩽ r n inf j ⩾ n   A j ( k ) , ∀ n ⩾ N , {:(5)(r_(n))/(s u p_(j rarr n)A_(j)(k)) <= R_(n+k) <= (r_(n))/(i n f_(j >= n)A_(j)(k))","quad AA n >= N",":}\begin{equation*} \frac{r_{n}}{\sup _{j \rightarrow n} A_{j}(k)} \leqslant R_{n+k} \leqslant \frac{r_{n}}{\inf _{j \geqslant n} A_{j}(k)}, \quad \forall n \geqslant \mathrm{~N}, \tag{5} \end{equation*}(5)rnsupj→nAj(k)⩽Rn+k⩽rninfj⩾nAj(k),∀n⩾ N,
where R m = S − S m R m = S − S m R_(m)=S-S_(m)R_{m}=S-S_{m}Rm=S−Sm.
In the same way, if condition (4) is satisfied, we obtain :
(6) r n sup j ⩾ n A j ( k ) ⩾ R n + k ⩾ r n inf j ⩾ n A j ( k ) , ∀ n ⩾ N . (6) r n sup j ⩾ n   A j ( k ) ⩾ R n + k ⩾ r n inf j ⩾ n   A j ( k ) , ∀ n ⩾ N . {:(6)(r_(n))/(s u p_(j >= n)A_(j)(k)) >= R_(n+k) >= (r_(n))/(i n f_(j >= n)A_(j)(k))","AA n >= N.:}\begin{equation*} \frac{r_{n}}{\sup _{j \geqslant n} A_{j}(k)} \geqslant R_{n+k} \geqslant \frac{r_{n}}{\inf _{j \geqslant n} A_{j}(k)}, \forall n \geqslant N . \tag{6} \end{equation*}(6)rnsupj⩾nAj(k)⩾Rn+k⩾rninfj⩾nAj(k),∀n⩾N.
Then, if ( A n ( k ) ) n A n ( k ) n (A_(n)(k))_(n)\left(A_{n}(k)\right)_{n}(An(k))n converges, we have
lim n → ∞ R n + k ¯ r n / A ˙ n ( k ) = 1 , lim n → ∞   R n + k ¯ r n / A ˙ n ( k ) = 1 , lim_(n rarr oo)(R_(n+ bar(k)))/(r_(n)//A^(˙)_(n)(k))=1,\lim _{n \rightarrow \infty} \frac{R_{n+\bar{k}}}{r_{n} / \dot{A}_{n}(k)}=1,limn→∞Rn+k¯rn/A˙n(k)=1,
which means that D n = r n − k A n − k ( k ) , n ∈ N ( k ∈ N D n = r n − k A n − k ( k ) , n ∈ N ( k ∈ N D_(n)=(r_(n-k))/(A_(n-k)(k)),n inN(k inND_{n}=\frac{r_{n-k}}{A_{n-k}(k)}, n \in \mathbb{N}(k \in \mathbb{N}Dn=rn−kAn−k(k),n∈N(k∈N, fixed) is a "perfect estimation of the error" of ( S n S n S_(n)S_{n}Sn ) and we get the following Theorem:
Theorem 1 [2]. Let ( S n ) S n (S_(n))\left(S_{n}\right)(Sn) be a monotone sequence and ( x n ) x n (x_(n))\left(x_{n}\right)(xn) an auxiliary one, such that:
a) lim n → ∞ x n = x lim n → ∞   x n = x lim_(n rarr oo)x_(n)=x\lim _{n \rightarrow \infty} x_{n}=xlimn→∞xn=x, where x x xxx is a finite number
b) ( A n ( k ) ) A n ( k ) (A_(n)(k))\left(A_{n}(k)\right)(An(k)) is convergent and lim n → ∞ A n ( k ) ≠ 0 lim n → ∞   A n ( k ) ≠ 0 lim_(n rarr oo)A_(n)(k)!=0\lim _{n \rightarrow \infty} A_{n}(k) \neq 0limn→∞An(k)≠0, with ( A n ( k ) ) A n ( k ) (A_(n)(k))\left(A_{n}(k)\right)(An(k)) defined by (2).
Then ( S n S n S_(n)S_{n}Sn ) converges and the transformation
T n = S n + x − x n − s A n − s ( k ) , n ∈ N ( k ∈ N , fixed ) T n = S n + x − x n − s A n − s ( k ) , n ∈ N ( k ∈ N ,  fixed  ) T_(n)=S_(n)+(x-x_(n-s))/(A_(n-s)(k)),quad n inNquad(k inN," fixed ")T_{n}=\mathbb{S}_{n}+\frac{x-x_{n-s}}{A_{n-s}(k)}, \quad n \in \mathbb{N} \quad(k \in \mathbb{N}, \text { fixed })Tn=Sn+x−xn−sAn−s(k),n∈N(k∈N, fixed )
accelerates the convergence of ( S n S n S_(n)S_{n}Sn ).
In the following we shall make some observations concerning this result.

2. AN ACCELERATION METHOD

Inequalities (5) and (6) can be written in the form:
1 sup j ⩾ n A j ( k ) ⩽ R n + k r n ⩽ 1 inf j ⩾ n A j ( k ) , ∀ n ⩾ N , 1 sup j ⩾ n   A j ( k ) ⩽ R n + k r n ⩽ 1 inf j ⩾ n   A j ( k ) , ∀ n ⩾ N , (1)/(s u p_(j >= n)A_(j)(k)) <= (R_(n+k))/(r_(n)) <= (1)/(i n f_(j >= n)A_(j)(k)),AA n >= N,\frac{1}{\sup _{j \geqslant n} A_{j}(k)} \leqslant \frac{R_{n+k}}{r_{n}} \leqslant \frac{1}{\inf _{j \geqslant n} A_{j}(k)}, \forall n \geqslant N,1supj⩾nAj(k)⩽Rn+krn⩽1infj⩾nAj(k),∀n⩾N,
respectively
1 sup j ⩾ n A j ( k ) ⩾ R n + k r n ⩾ 1 inf j ⩾ n A j ( k ) , ∀ n ⩾ N . 1 sup j ⩾ n   A j ( k ) ⩾ R n + k r n ⩾ 1 inf j ⩾ n   A j ( k ) , ∀ n ⩾ N . (1)/(s u p_(j >= n)A_(j)(k)) >= (R_(n+k))/(r_(n)) >= (1)/(i n f_(j >= n)A_(j)(k)),AA n >= N.\frac{1}{\sup _{j \geqslant n} A_{j}(k)} \geqslant \frac{R_{n+k}}{r_{n}} \geqslant \frac{1}{\inf _{j \geqslant n} A_{j}(k)}, \forall n \geqslant N .1supj⩾nAj(k)⩾Rn+krn⩾1infj⩾nAj(k),∀n⩾N.
If lim n → ∞ A n ( k ) = μ lim n → ∞   A n ( k ) = μ lim_(n rarr oo)A_(n)(k)=mu\lim _{n \rightarrow \infty} A_{n}(k)=\mulimn→∞An(k)=μ, where μ ≠ 0 μ ≠ 0 mu!=0\mu \neq 0μ≠0 is a finite number, we have :
lim n → ∞ R n + k r n = 1 μ , lim n → ∞   R n + k r n = 1 μ , lim_(n rarr oo)(R_(n+k))/(r_(n))=(1)/(mu),\lim _{n \rightarrow \infty} \frac{R_{n+k}}{r_{n}}=\frac{1}{\mu},limn→∞Rn+krn=1μ,
so
lim n → ∞ R n + k r n / μ = 1 lim n → ∞   R n + k r n / μ = 1 lim_(n rarr oo)(R_(n+k))/(r_(n)//mu)=1\lim _{n \rightarrow \infty} \frac{R_{n+k}}{r_{n} / \mu}=1limn→∞Rn+krn/μ=1
which means that D n = r n − k μ , n ∈ N ( k ∈ N D n = r n − k μ , n ∈ N ( k ∈ N D_(n)=(r_(n-k))/(mu),n inN(k inND_{n}=\frac{r_{n-k}}{\mu}, n \in \mathbb{N}(k \in \mathbb{N}Dn=rn−kμ,n∈N(k∈N, fixed) is a "perfect estimation of the error" of ( S n S n S_(n)S_{n}Sn ). It follows that, in the conditions of Theorem 1, the sequence ( T n T n T_(n)T_{n}Tn ) given by
T n = S n + x − x n − k μ , n ∈ N ( k ∈ N , fixed ) T n = S n + x − x n − k μ , n ∈ N ( k ∈ N ,  fixed  ) T_(n)=S_(n)+(x-x_(n-k))/(mu),quad n inN(k inN," fixed ")T_{n}=S_{n}+\frac{x-x_{n-k}}{\mu}, \quad n \in \mathbb{N}(k \in \mathbb{N}, \text { fixed })Tn=Sn+x−xn−kμ,n∈N(k∈N, fixed )
accelerates the convergence of ( S n S n S_(n)S_{n}Sn ).
Starting from the results in [3], concerning the accelerating convergence of series with positive terms, the following problem arises :
Determine the relations between two sequences ( x n ( 1 ) ) x n ( 1 ) (x_(n)^((1)))\left(x_{n}^{(1)}\right)(xn(1)) and ( x n ( 2 ) ) x n ( 2 ) (x_(n)^((2)))\left(x_{n}^{(2)}\right)(xn(2)) so that the difference between R n = S − S n R n = S − S n R_(n)=S-S_(n)R_{n}=S-S_{n}Rn=S−Sn and D n ( 1 ) = r n − k ( 1 ) n D n ( 1 ) = r n − k ( 1 ) n D_(n)^((1))=(r_(n-k)^((1)))/(n)D_{n}^{(1)}=\frac{r_{n-k}^{(1)}}{n}Dn(1)=rn−k(1)n (the asympto. tic expression of R n R n R_(n)R_{n}Rn obtained using ( x n ( 1 ) x n ( 1 ) x_(n)^((1))x_{n}^{(1)}xn(1) )) tend to zero faster than the difference between R n R n R_(n)R_{n}Rn and D n ( 2 ) = r n − k ( 2 ) μ D n ( 2 ) = r n − k ( 2 ) μ D_(n)^((2))=(r_(n-k)^((2)))/(mu)D_{n}^{(2)}=\frac{r_{n-k}^{(2)}}{\mu}Dn(2)=rn−k(2)μ (the asymptotic expresion of R n R n R_(n)R_{n}Rn obtained using ( x n ( 2 ) ) x n ( 2 ) {:x_(n)^((2)))\left.x_{n}^{(2)}\right)xn(2)) ), where
r n − k ( i ) = x ( i ) − x n − k ( i ) , x ( i ) = lim x n ( i ) , i = 1 , 2 . r n − k ( i ) = x ( i ) − x n − k ( i ) , x ( i ) = lim x n ( i ) , i = 1 , 2 . r_(n-k)^((i))=x^((i))-x_(n-k)^((i)),quadx^((i))=limx_(n)^((i)),quad i=1,2.r_{n-k}^{(i)}=x^{(i)}-x_{n-k}^{(i)}, \quad x^{(i)}=\lim x_{n}^{(i)}, \quad i=1,2 .rn−k(i)=x(i)−xn−k(i),x(i)=limxn(i),i=1,2.
In this respect we present the following results.
Lemma 1. Let ( S n ) S n (S_(n))\left(S_{n}\right)(Sn) be a monotone sequence and ( x n ( 1 ) ) , ( x n ( 2 ) ) x n ( 1 ) , x n ( 2 ) (x_(n)^((1))),(x_(n)^((2)))\left(x_{n}^{(1)}\right),\left(x_{n}^{(2)}\right)(xn(1)),(xn(2)) auxiliary sequences such that:
a)
lim n → ∞ w n ( i ) = x ( i ) , i = 1 , 2 lim n → ∞   w n ( i ) = x ( i ) , i = 1 , 2 lim_(n rarr oo)w_(n)^((i))=x^((i)),i=1,2\lim _{n \rightarrow \infty} w_{n}^{(i)}=x^{(i)}, i=1,2limn→∞wn(i)=x(i),i=1,2
where x ( i ) x ( i ) x^((i))x^{(i)}x(i) are finite numbers,
b)
lim n → ∞ Δ x n ( i ) Δ S m + k = μ ≠ 0 , i = 1 , 2 lim n → ∞   Δ x n ( i ) Δ S m + k = μ ≠ 0 , i = 1 , 2 lim_(n rarr oo)(Deltax_(n)^((i)))/(DeltaS_(m+k))=mu!=0,i=1,2\lim _{n \rightarrow \infty} \frac{\Delta x_{n}^{(i)}}{\Delta S_{m+k}}=\mu \neq 0, i=1,2limn→∞Δxn(i)ΔSm+k=μ≠0,i=1,2
Then ( x n ( 1 ) x n ( 1 ) x_(n)^((1))x_{n}^{(1)}xn(1) ) and ( ω n ( 2 ) ω n ( 2 ) omega_(n)^((2))\omega_{n}^{(2)}ωn(2) ) satisfy:
lim n → ∞ x ( 2 ) − x n − k ( 2 ) x ( 1 ) − x n − k ( 1 ) = 1 lim n → ∞   x ( 2 ) − x n − k ( 2 ) x ( 1 ) − x n − k ( 1 ) = 1 lim_(n rarr oo)(x^((2))-x_(n-k)^((2)))/(x^((1))-x_(n-k)^((1)))=1\lim _{n \rightarrow \infty} \frac{x^{(2)}-x_{n-k}^{(2)}}{x^{(1)}-x_{n-k}^{(1)}}=1limn→∞x(2)−xn−k(2)x(1)−xn−k(1)=1
Proof. From Theorem 1, we have that ( S n ) S n (S_(n))\left(S_{n}\right)(Sn) is convergent and D n ( 1 ) == x ( 1 ) − r n − k ( 1 ) μ D n ( 1 ) == x ( 1 ) − r n − k ( 1 ) μ D_(n)^((1))==(x^((1))-r_(n-k)^((1)))/(mu)D_{n}^{(1)}= =\frac{x^{(1)}-\mathrm{r}_{n-k}^{(1)}}{\mu}Dn(1)==x(1)−rn−k(1)μ respectively D n ( 2 ) = x ( 2 ) − x n − k ( 2 ) μ D n ( 2 ) = x ( 2 ) − x n − k ( 2 ) μ D_(n)^((2))=(x^((2))-x_(n-k)^((2)))/(mu)D_{n}^{(2)}=\frac{x^{(2)}-x_{n-k}^{(2)}}{\mu}Dn(2)=x(2)−xn−k(2)μ are asymptotic expressions of the rest R n = S − S n R n = S − S n R_(n)=S-S_(n)R_{n}=S-S_{n}Rn=S−Sn. So, lim n → ∞ D n ( 4 ) R n = 1 , i = 1 , 2 lim n → ∞   D n ( 4 ) R n = 1 , i = 1 , 2 lim_(n rarr oo)(D_(n)^((4)))/(R_(n))=1,i=1,2\lim _{n \rightarrow \infty} \frac{D_{n}^{(4)}}{R_{n}}=1, i=1,2limn→∞Dn(4)Rn=1,i=1,2 imply lim n → ∞ D n ( 2 ) D n ( 1 ) == 1 lim n → ∞   D n ( 2 ) D n ( 1 ) == 1 lim_(n rarr oo)(D_(n)^((2)))/(D_(n)^((1)))==1\lim _{n \rightarrow \infty} \frac{D_{n}^{(2)}}{D_{n}^{(1)}}= =1limn→∞Dn(2)Dn(1)==1, which means that lim n → ∞ x ( 2 ) − x n − k ( 2 ) x ( 1 ) − x n − k ( 1 ) = 1 lim n → ∞   x ( 2 ) − x n − k ( 2 ) x ( 1 ) − x n − k ( 1 ) = 1 lim_(n rarr oo)(x^((2))-x_(n-k)^((2)))/(x^((1))-x_(n-k)^((1)))=1\lim _{n \rightarrow \infty} \frac{x^{(2)}-x_{n-k}^{(2)}}{x^{(1)}-x_{n-k}^{(1)}}=1limn→∞x(2)−xn−k(2)x(1)−xn−k(1)=1.
LEMMA 2. Let ( S n ) , ( x n ( 1 ) ) S n , x n ( 1 ) (S_(n)),(x_(n)^((1)))\left(S_{n}\right),\left(x_{n}^{(1)}\right)(Sn),(xn(1)) and ( x n ( 2 ) ) x n ( 2 ) (x_(n)^((2)))\left(x_{n}^{(2)}\right)(xn(2)) sequences that satisfy the conditions of Lemma 1.
Let ( ε n ( 1 ) ) ε n ( 1 ) (epsi_(n)^((1)))\left(\varepsilon_{n}^{(1)}\right)(εn(1)) and ( ε n ( 2 ) ) ε n ( 2 ) (epsi_(n)^((2)))\left(\varepsilon_{n}^{(2)}\right)(εn(2)) be two sequences such that:
a) Δ x n − k ( i ) Δ S n − μ = ε n ( i ) Δ S n , i = 1 , 2 Δ x n − k ( i ) Δ S n − μ = ε n ( i ) Δ S n , i = 1 , 2 (Deltax_(n-k)^((i)))/(DeltaS_(n))-mu=(epsi_(n)^((i)))/(DeltaS_(n)),quad i=1,2\frac{\Delta x_{n-k}^{(i)}}{\Delta S_{n}}-\mu=\frac{\varepsilon_{n}^{(i)}}{\Delta S_{n}}, \quad i=1,2Δxn−k(i)ΔSn−μ=εn(i)ΔSn,i=1,2
b) ε n ( 1 ) ≠ 0 , ε n ( 2 ) = 0 , ∀ n ∈ N ε n ( 1 ) ≠ 0 , ε n ( 2 ) = 0 , ∀ n ∈ N epsi_(n)^((1))!=0,epsi_(n)^((2))=0,AA n inN\varepsilon_{n}^{(1)} \neq 0, \varepsilon_{n}^{(2)}=0, \forall n \in \mathbb{N}εn(1)≠0,εn(2)=0,∀n∈N,
c) ( ε n ( 1 ) ) ε n ( 1 ) (epsi_(n)^((1)))\left(\varepsilon_{n}^{(1)}\right)(εn(1)) keeps a constant sign beginning with some sufficiently large index n n nnn,
d) lim n → ∞ ε n ( 2 ) ε n ( 1 ) = 0 lim n → ∞   ε n ( 2 ) ε n ( 1 ) = 0 quadlim_(n rarr oo)(epsi_(n)^((2)))/(epsi_(n)^((1)))=0\quad \lim _{n \rightarrow \infty} \frac{\varepsilon_{n}^{(2)}}{\varepsilon_{n}^{(1)}}=0limn→∞εn(2)εn(1)=0,
then the asymptotic expressions D n ( 1 ) = x ( 1 ) − x n − k ( 1 ) μ D n ( 1 ) = x ( 1 ) − x n − k ( 1 ) μ D_(n)^((1))=(x^((1))-x_(n-k)^((1)))/(mu)D_{n}^{(1)}=\frac{x^{(1)}-x_{n-k}^{(1)}}{\mu}Dn(1)=x(1)−xn−k(1)μ and D n ( 2 ) = x ( 2 ) − x n − k ( 2 ) μ D n ( 2 ) = x ( 2 ) − x n − k ( 2 ) μ D_(n)^((2))=(x^((2))-x_(n-k)^((2)))/(mu)D_{n}^{(2)}=\frac{x^{(2)}-x_{n-k}^{(2)}}{\mu}Dn(2)=x(2)−xn−k(2)μ satisfy the relation:
lim n → ∞ D n ( 2 ) − R n D n ( 1 ) − R n = 0 lim n → ∞   D n ( 2 ) − R n D n ( 1 ) − R n = 0 lim_(n rarr oo)(D_(n)^((2))-R_(n))/(D_(n)^((1))-R_(n))=0\lim _{n \rightarrow \infty} \frac{D_{n}^{(2)}-R_{n}}{D_{n}^{(1)}-R_{n}}=0limn→∞Dn(2)−RnDn(1)−Rn=0
that is D n ( 2 ) D n ( 2 ) D_(n)^((2))D_{n}^{(2)}Dn(2) is an approximation for the rest better than D n ( 1 ) D n ( 1 ) D_(n)^((1))D_{n}^{(1)}Dn(1).
Proof. In the conditions imposed on the sequences ( S n ) , ( x n ( 1 ) ) S n , x n ( 1 ) (S_(n)),(x_(n)^((1)))\left(S_{n}\right),\left(x_{n}^{(1)}\right)(Sn),(xn(1)) and ( x n ( 2 ) x n ( 2 ) x_(n)^((2))x_{n}^{(2)}xn(2) ), we have that ( S n S n S_(n)S_{n}Sn ) is convergent.
Suppose now that ε n ( 1 ) > 0 ε n ( 1 ) > 0 epsi_(n)^((1)) > 0\varepsilon_{n}^{(1)}>0εn(1)>0 for n n nnn sufficiently large
Condition a) can be written as
x n − k + 1 ( i ) − x n − k ( i ) = μ ( S n + 1 − S n ) + ε n ( i ) , i = 1 , 2 x n − k + 1 ( i ) − x n − k ( i ) = μ S n + 1 − S n + ε n ( i ) , i = 1 , 2 x_(n-k+1)^((i))-x_(n-k)^((i))=mu(S_(n+1)-S_(n))+epsi_(n)^((i)),i=1,2x_{n-k+1}^{(i)}-x_{n-k}^{(i)}=\mu\left(S_{n+1}-S_{n}\right)+\varepsilon_{n}^{(i)}, i=1,2xn−k+1(i)−xn−k(i)=μ(Sn+1−Sn)+εn(i),i=1,2
and adding from n n nnn to ∞ ∞ oo\infty∞, we obtain:
x ( 1 ) − x n − k ( 1 ) = μ ( S − S n ) + ∑ v = n ∞ ϵ v ( 1 ) (7) x ( 2 ) − x n − k ( 2 ) = μ ( S − S n ) + ∑ v = n ∞ ϵ v ( 2 ) x ( 1 ) − x n − k ( 1 ) = μ S − S n + ∑ v = n ∞   ϵ v ( 1 ) (7) x ( 2 ) − x n − k ( 2 ) = μ S − S n + ∑ v = n ∞   ϵ v ( 2 ) {:[x^((1))-x_(n-k)^((1))=mu(S-S_(n))+sum_(v=n)^(oo)epsilon_(v)^((1))],[(7)x^((2))-x_(n-k)^((2))=mu(S-S_(n))+sum_(v=n)^(oo)epsilon_(v)^((2))]:}\begin{align*} x^{(1)}-x_{n-k}^{(1)} & =\mu\left(S-S_{n}\right)+\sum_{v=n}^{\infty} \epsilon_{v}^{(1)} \\ x^{(2)}-x_{n-k}^{(2)} & =\mu\left(S-S_{n}\right)+\sum_{v=n}^{\infty} \epsilon_{v}^{(2)} \tag{7} \end{align*}x(1)−xn−k(1)=μ(S−Sn)+∑v=n∞ϵv(1)(7)x(2)−xn−k(2)=μ(S−Sn)+∑v=n∞ϵv(2)
The relation lim n → ∞ ε n ( 2 ) ε n ( 1 ) = 0 lim n → ∞   ε n ( 2 ) ε n ( 1 ) = 0 lim_(n rarr oo)(epsi_(n)^((2)))/(epsi_(n)^((1)))=0\lim _{n \rightarrow \infty} \frac{\varepsilon_{n}^{(2)}}{\varepsilon_{n}^{(1)}}=0limn→∞εn(2)εn(1)=0 can be written as :
(8) − α < ε n ( 2 ) ε n ( 1 ) < α ( n > N α ) , (8) − α < ε n ( 2 ) ε n ( 1 ) < α n > N α , {:(8)-alpha < (epsi_(n)^((2)))/(epsi_(n)^((1))) < alpha(n > N_(alpha))",":}\begin{equation*} -\alpha<\frac{\varepsilon_{n}^{(2)}}{\varepsilon_{n}^{(1)}}<\alpha\left(n>N_{\alpha}\right), \tag{8} \end{equation*}(8)−α<εn(2)εn(1)<α(n>Nα),
where α α alpha\alphaα is an arbitrary positive number and N α N α N_(alpha)N_{\alpha}Nα is a positive integer associated to α α alpha\alphaα.
By condition c), from (8) we have the following inequalities:
− α ∑ v = n ∞ ε v ( 1 ) < ∑ v = n ∞ ε v ( 2 ) < α ∑ v = n ∞ ε v ( 1 ) , ( n > N α ) − α ∑ v = n ∞   ε v ( 1 ) < ∑ v = n ∞   ε v ( 2 ) < α ∑ v = n ∞   ε v ( 1 ) , n > N α -alphasum_(v=n)^(oo)epsi_(v)^((1)) < sum_(v=n)^(oo)epsi_(v)^((2)) < alphasum_(v=n)^(oo)epsi_(v)^((1)),quad(n > N_(alpha))-\alpha \sum_{v=n}^{\infty} \varepsilon_{v}^{(1)}<\sum_{v=n}^{\infty} \varepsilon_{v}^{(2)}<\alpha \sum_{v=n}^{\infty} \varepsilon_{v}^{(1)}, \quad\left(n>N_{\alpha}\right)−α∑v=n∞εv(1)<∑v=n∞εv(2)<α∑v=n∞εv(1),(n>Nα)
that is,
(9) − α < ∑ v = n ∞ ϵ v ( 2 ) ∑ v = n ∞ ϵ v ( 1 ) < α ( n > N α ) . (9) − α < ∑ v = n ∞   ϵ v ( 2 ) ∑ v = n ∞   ϵ v ( 1 ) < α n > N α . {:(9)-alpha < (sum_(v=n)^(oo)epsilon_(v)^((2)))/(sum_(v=n)^(oo)epsilon_(v)^((1))) < alphaquad(n > N_(alpha)).:}\begin{equation*} -\alpha<\frac{\sum_{v=n}^{\infty} \epsilon_{v}^{(2)}}{\sum_{v=n}^{\infty} \epsilon_{v}^{(1)}}<\alpha \quad\left(n>N_{\alpha}\right) . \tag{9} \end{equation*}(9)−α<∑v=n∞ϵv(2)∑v=n∞ϵv(1)<α(n>Nα).
Taking into account (7), inequalities (9) are equivalent to the relation
(10) lim n → ∞ x ( 2 ) − x n − k ( 2 ) μ − ( S − S n ) x ( 1 ) − x n − k ( 1 ) μ − ( S − S n ) = 0 (10) lim n → ∞   x ( 2 ) − x n − k ( 2 ) μ − S − S n x ( 1 ) − x n − k ( 1 ) μ − S − S n = 0 {:(10)lim_(n rarr oo)((x^((2))-x_(n-k)^((2)))/(mu)-(S-S_(n)))/((x^((1))-x_(n-k)^((1)))/(mu)-(S-S_(n)))=0:}\begin{equation*} \lim _{n \rightarrow \infty} \frac{\frac{x^{(2)}-x_{n-k}^{(2)}}{\mu}-\left(S-S_{n}\right)}{\frac{x^{(1)}-x_{n-k}^{(1)}}{\mu}-\left(S-S_{n}\right)}=0 \tag{10} \end{equation*}(10)limn→∞x(2)−xn−k(2)μ−(S−Sn)x(1)−xn−k(1)μ−(S−Sn)=0
Remarks. 1) If, beginning with an index n n nnn sufficiently large, ε n ( 1 ) ε n ( 1 ) epsi_(n)^((1))\varepsilon_{n}^{(1)}εn(1) is negative, then the reasoning is not essentialy modified.
2) The solution (10) is equivalent to:
The transformation
P n = S n + x ( 2 ) − x n − ( 2 ) k μ , n ∈ N ( k ∈ N , f i x e d ) P n = S n + x ( 2 ) − x n − ( 2 ) k μ , n ∈ N ( k ∈ N , f i x e d ) P_(n)=S_(n)+(x^((2))-x_(n-)^((2))k)/(mu),n inN(k inN,fixed)P_{n}=S_{n}+\frac{x^{(2)}-x_{n-}^{(2)} k}{\mu}, n \in \mathbb{N}(k \in \mathbb{N}, f i x e d)Pn=Sn+x(2)−xn−(2)kμ,n∈N(k∈N,fixed)
accelerates the convergence of the transformation,
T n = S n + x ( 1 ) − x n − k ( 1 ) μ , n ∈ N ( k ∈ N , fixed ) T n = S n + x ( 1 ) − x n − k ( 1 ) μ , n ∈ N ( k ∈ N ,  fixed  ) T_(n)=S_(n)+(x^((1))-x_(n-k)^((1)))/(mu),n in N(k inN," fixed ")T_{n}=S_{n}+\frac{x^{(1)}-x_{n-k}^{(1)}}{\mu}, n \in N(k \in \mathbb{N}, \text { fixed })Tn=Sn+x(1)−xn−k(1)μ,n∈N(k∈N, fixed )
The method of acceleration. Suppose ( S n S n S_(n)S_{n}Sn ) is a monotone sequence, ( x n ) x n (x_(n))\left(x_{n}\right)(xn) is a convergent sequence, lim n → ∞ x n = x lim n → ∞   x n = x lim_(n rarr oo)x_(n)=x\lim _{n \rightarrow \infty} x_{n}=xlimn→∞xn=x, for which :
lim n → ∞ Δ x n Δ S n + k = μ ≠ 0 . lim n → ∞   Δ x n Δ S n + k = μ ≠ 0 . lim_(n rarr oo)(Deltax_(n))/(DeltaS_(n+k))=mu!=0.\lim _{n \rightarrow \infty} \frac{\Delta x_{n}}{\Delta S_{n+k}}=\mu \neq 0 .limn→∞ΔxnΔSn+k=μ≠0.
If the expression x n − k + 1 − x n − k − μ ( S n + 1 − S n ) x n − k + 1 − x n − k − μ S n + 1 − S n x_(n_(-)k+1)-x_(n_(-)k)-mu(S_(n_(+1))-S_(n))x_{n_{-} k+1}-x_{n_{-} k}-\mu\left(S_{n_{+1}}-S_{n}\right)xn−k+1−xn−k−μ(Sn+1−Sn) keeps a constant sign for large enough indexes, then we are looking for a sequence ( b n b n b_(n)b_{n}bn ) such that:
(11) lim n → ∞ b n = 1 (12) lim n → ∞ x n − k + 1 b n − k + 1 − x n − k b n − k − μ ( S n + 1 − S n ) x n − k + 1 − x n − k − μ ( S n + 1 − S n ) = 0 (11) lim n → ∞   b n = 1 (12) lim n → ∞   x n − k + 1 b n − k + 1 − x n − k b n − k − μ S n + 1 − S n x n − k + 1 − x n − k − μ S n + 1 − S n = 0 {:[(11)lim_(n rarr oo)b_(n)=1],[(12)lim_(n rarr oo)(x_(n-k+1)b_(n-k+1)-x_(n-k)b_(n-k)-mu(S_(n+1)-S_(n)))/(x_(n-k+1)-x_(n-k)-mu(S_(n+1)-S_(n)))=0]:}\begin{align*} & \lim _{n \rightarrow \infty} b_{n}=1 \tag{11}\\ & \lim _{n \rightarrow \infty} \frac{x_{n-k+1} b_{n-k+1}-x_{n-k} b_{n-k}-\mu\left(S_{n+1}-S_{n}\right)}{x_{n-k+1}-x_{n-k}-\mu\left(S_{n+1}-S_{n}\right)}=0 \tag{12} \end{align*}(11)limn→∞bn=1(12)limn→∞xn−k+1bn−k+1−xn−kbn−k−μ(Sn+1−Sn)xn−k+1−xn−k−μ(Sn+1−Sn)=0
In these conditions, the sequence
P n = S n + x − x n − k b n − k μ , n ∈ N ( k ∈ N , fixed ) P n = S n + x − x n − k b n − k μ , n ∈ N ( k ∈ N ,  fixed  ) P_(n)=S_(n)+(x-x_(n-k)b_(n-k))/(mu),n inN(k inN," fixed ")P_{n}=S_{n}+\frac{x-x_{n-k} b_{n-k}}{\mu}, n \in \mathbb{N}(k \in \mathbb{N}, \text { fixed })Pn=Sn+x−xn−kbn−kμ,n∈N(k∈N, fixed )
accelerates the convergence of ( T n T n T_(n)T_{n}Tn ), where
T n = S n + x − x n − k μ , n ∈ N ( k ∈ N , fixed ) T n = S n + x − x n − k μ , n ∈ N ( k ∈ N ,  fixed  ) T_(n)=S_(n)+(x-x_(n-k))/(mu),n inNquad(k inN," fixed ")T_{n}=S_{n}+\frac{x-x_{n-k}}{\mu}, n \in \mathbb{N} \quad(k \in \mathbb{N}, \text { fixed })Tn=Sn+x−xn−kμ,n∈N(k∈N, fixed )
which, in its turn (cf. Theorem 1) accelerates the convergence of ( S n S n S_(n)S_{n}Sn ).
We say that ( P n P n P_(n)P_{n}Pn ) accelerates the convergence of ( S n S n S_(n)S_{n}Sn ) better than ( T n T n T_(n)T_{n}Tn ).
Remarks. The sequence ( b n b n b_(n)b_{n}bn ) may be taken, for example, in the following form :
b n = 1 + β n , ∀ n ∈ N , β ∈ R ∗ , b n = 1 + β n , ∀ n ∈ N , β ∈ R ∗ , b_(n)=1+(beta )/(n),quad AA n inN,beta inR^(**),b_{n}=1+\frac{\beta}{n}, \quad \forall n \in \mathbb{N}, \beta \in R^{*},bn=1+βn,∀n∈N,β∈R∗,
in order that (12) hold.
Numerical example 1
Let S n = ∑ k = 1 n 1 k 2 , n ⩾ 1 , lim n → ∞ S n = S = π 2 6 ≈ 1.6449341 S n = ∑ k = 1 n   1 k 2 , n ⩾ 1 , lim n → ∞   S n = S = π 2 6 ≈ 1.6449341 S_(n)=sum_(k=1)^(n)(1)/(k^(2)),n >= 1,lim_(n rarr oo)S_(n)=S=(pi^(2))/(6)~~1.6449341S_{n}=\sum_{k=1}^{n} \frac{1}{k^{2}}, n \geqslant 1, \lim _{n \rightarrow \infty} S_{n}=S=\frac{\pi^{2}}{6} \approx 1.6449341Sn=∑k=1n1k2,n⩾1,limn→∞Sn=S=π26≈1.6449341
Consider ( x n ) x n (x_(n))\left(x_{n}\right)(xn) defined by :
x n = n Δ S n , ∀ n ∈ N , x n = n Δ S n , ∀ n ∈ N , x_(n)=n DeltaS_(n),AA n inN,x_{n}=n \Delta S_{n}, \forall n \in \mathbb{N},xn=nΔSn,∀n∈N,
that is,
x n = n ( n + 1 ) 2 , ∀ n ∈ N . x n = n ( n + 1 ) 2 , ∀ n ∈ N . x_(n)=(n)/((n+1)^(2)),quad AA n inN.x_{n}=\frac{n}{(n+1)^{2}}, \quad \forall n \in \mathbb{N} .xn=n(n+1)2,∀n∈N.
We have that lim n → ∞ x n = 0 lim n → ∞   x n = 0 lim_(n rarr oo)x_(n)=0\lim _{n \rightarrow \infty} x_{n}=0limn→∞xn=0.
In these conditions A n ( 0 ) = Δ x n Δ S n = − n 2 − n + 1 n 2 + 4 n + 1 , ∀ n ∈ N A n ( 0 ) = Δ x n Δ S n = − n 2 − n + 1 n 2 + 4 n + 1 , ∀ n ∈ N A_(n)(0)=(Deltax_(n))/(DeltaS_(n))=(-n^(2)-n+1)/(n^(2)+4n+1),quad AA n inNA_{n}(0)=\frac{\Delta x_{n}}{\Delta S_{n}}=\frac{-n^{2}-n+1}{n^{2}+4 n+1}, \quad \forall n \in \mathbb{N}An(0)=ΔxnΔSn=−n2−n+1n2+4n+1,∀n∈N, and lim n → ∞ − Δ x n Δ S n = − 1 ≠ 0 lim n → ∞   − Δ x n Δ S n = − 1 ≠ 0 lim_(n rarr oo)-(Deltax_(n))/(DeltaS_(n))=-1!=0\lim _{n \rightarrow \infty}-\frac{\Delta x_{n}}{\Delta S_{n}}=-1 \neq 0limn→∞−ΔxnΔSn=−1≠0.
We are looking for ( b n ) b n (b_(n))\left(b_{n}\right)(bn) in the form b n = 1 + β n ( β ∈ R ∗ ) b n = 1 + β n β ∈ R ∗ b_(n)=1+(beta )/(n)quad(beta inR^(**))b_{n}=1+\frac{\beta}{n} \quad\left(\beta \in R^{*}\right)bn=1+βn(β∈R∗).
From (12) we get β = 3 2 β = 3 2 beta=(3)/(2)\beta=\frac{3}{2}β=32.
The conditions being fultilled, as shown in table 1
P n = S n + x − x n b n μ , n ∈ N P n = S n + x − x n b n μ , n ∈ N P_(n)=S_(n)+(x-x_(n)b_(n))/(mu),n inNP_{n}=S_{n}+\frac{x-x_{n} b_{n}}{\mu}, n \in \mathbb{N}Pn=Sn+x−xnbnμ,n∈N
accelerates the convergence of ( S n S n S_(n)S_{n}Sn ) better than,
T n = S n + x − x n μ , n ∈ N T n = S n + x − x n μ , n ∈ N T_(n)=S_(n)+(x-x_(n))/(mu),n inNT_{n}=S_{n}+\frac{x-x_{n}}{\mu}, n \in \mathbb{N}Tn=Sn+x−xnμ,n∈N
Table 1
n n nnn S 12 S 12 S_(12)S_{12}S12 T 16 T 16 T_(16)T_{16}T16 P n P n P_(n)P_{n}Pn
1 1.0000000 1.2500000 1.6250000
2 1.2500000 1.4722222 1.6388889
3 1.3611111 1.5486111 1.6423611
4 1.4236111 2.5836111 1. 6436111
5 1.4636111 1.6025000 1.6441667
6 1.4913889 1.6138379 1.6444501
7 1.5117971 1,6211721 1.6440006
8 1.5274221 1.6261875 1.6447060
9 1.5397677 1.6297677 1.6447677
10 1.5497677 1.6324124 1.6448001
11 1.5580322 1.6342211 1.6448377
12 1.5649766 1.6359826 1.6448583
13 1.5708938 1.6372203 1.6448734
14 1.5759958 1.6382181 1.6448847
15 1.5804403 1.6390340 1.6448934
16 1.5843465 1.6397099 1.6449002
17 1.5878067 1.6402759 1.6449055
n S_(12) T_(16) P_(n) 1 1.0000000 1.2500000 1.6250000 2 1.2500000 1.4722222 1.6388889 3 1.3611111 1.5486111 1.6423611 4 1.4236111 2.5836111 1. 6436111 5 1.4636111 1.6025000 1.6441667 6 1.4913889 1.6138379 1.6444501 7 1.5117971 1,6211721 1.6440006 8 1.5274221 1.6261875 1.6447060 9 1.5397677 1.6297677 1.6447677 10 1.5497677 1.6324124 1.6448001 11 1.5580322 1.6342211 1.6448377 12 1.5649766 1.6359826 1.6448583 13 1.5708938 1.6372203 1.6448734 14 1.5759958 1.6382181 1.6448847 15 1.5804403 1.6390340 1.6448934 16 1.5843465 1.6397099 1.6449002 17 1.5878067 1.6402759 1.6449055| $n$ | $S_{12}$ | $T_{16}$ | $P_{n}$ | | :--- | :--- | :--- | :--- | | 1 | 1.0000000 | 1.2500000 | 1.6250000 | | 2 | 1.2500000 | 1.4722222 | 1.6388889 | | 3 | 1.3611111 | 1.5486111 | 1.6423611 | | 4 | 1.4236111 | 2.5836111 | 1. 6436111 | | 5 | 1.4636111 | 1.6025000 | 1.6441667 | | 6 | 1.4913889 | 1.6138379 | 1.6444501 | | 7 | 1.5117971 | 1,6211721 | 1.6440006 | | 8 | 1.5274221 | 1.6261875 | 1.6447060 | | 9 | 1.5397677 | 1.6297677 | 1.6447677 | | 10 | 1.5497677 | 1.6324124 | 1.6448001 | | 11 | 1.5580322 | 1.6342211 | 1.6448377 | | 12 | 1.5649766 | 1.6359826 | 1.6448583 | | 13 | 1.5708938 | 1.6372203 | 1.6448734 | | 14 | 1.5759958 | 1.6382181 | 1.6448847 | | 15 | 1.5804403 | 1.6390340 | 1.6448934 | | 16 | 1.5843465 | 1.6397099 | 1.6449002 | | 17 | 1.5878067 | 1.6402759 | 1.6449055 |

4. ANOTHER FORMULATION FOR THEOREM :

Let ( S n S n S_(n)S_{n}Sn ) be a monotone sequence, ( a n a n a_(n)a_{n}an ) an auxiliary sequence such that ( Δ p − 1 a n Δ p − 1 a n Delta^(p-1)a_(n)\Delta^{p-1} a_{n}Δp−1an ) converges to the finite limit l l lll, where p ∈ N ∗ , Δ p − 1 a k n == Δ ( Δ p − 1 a n ) , Δ ∘ a n = a n p ∈ N ∗ , Δ p − 1 a k n == Δ Δ p − 1 a n , Δ ∘ a n = a n p inN^(**),Delta^(p-1)a_(kn)==Delta(Delta^(p-1)a_(n)),Delta^(@)a_(n)=a_(n)p \in \mathbb{N}^{*}, \Delta^{p-1} a_{k n}= =\Delta\left(\Delta^{p-1} a_{n}\right), \Delta^{\circ} a_{n}=a_{n}p∈N∗,Δp−1akn==Δ(Δp−1an),Δ∘an=an.
Define :
l n = l − Δ p − 1 a n B n ( p , k ) = Δ p a n Δ S n + k , n ∈ N ( k , p ∈ N , fixed ) l n = l − Δ p − 1 a n B n ( p , k ) = Δ p a n Δ S n + k , n ∈ N ( k , p ∈ N ,  fixed  ) {:[l_(n)=l-Delta^(p-1)a_(n)],[B_(n)(p","k)=(Delta^(p)a_(n))/(DeltaS_(n_(+k)))","n inN(k","p inN","" fixed ")]:}\begin{gathered} l_{n}=l-\Delta^{p-1} a_{n} \\ B_{n}(p, k)=\frac{\Delta^{p} a_{n}}{\Delta S_{n_{+k}}}, n \in \mathbb{N}(k, p \in \mathbb{N}, \text { fixed }) \end{gathered}ln=l−Δp−1anBn(p,k)=ΔpanΔSn+k,n∈N(k,p∈N, fixed )
The reasoning presented in Introduction, permits us the following test of convergence:
If
lim n → ∞ inf B n ( p , k ) > 0 lim n → ∞   inf B n ( p , k ) > 0 lim_(n rarr oo)i n fB_(n)(p,k) > 0\lim _{n \rightarrow \infty} \inf B_{n}(p, k)>0limn→∞infBn(p,k)>0
or
lim sup n → ∞ B n ( p , k ) < 0 lim sup n → ∞   B n ( p , k ) < 0 l i m   s u p_(n rarr oo)B_(n)(p,k) < 0\limsup _{n \rightarrow \infty} B_{n}(p, k)<0lim supn→∞Bn(p,k)<0
then ( S n S n S_(n)S_{n}Sn ) is convergent.
This test enables us to construct an estimation for the difference ( S − S n ) S − S n (S-S_(n))\left(S-S_{n}\right)(S−Sn) and the formulation of the following result:
Theorem 2. Let ( S n S n S_(n)S_{n}Sn ) be a monotone sequence and ( a n a n a_(n)a_{n}an ) an auxiliary one, such that:
a) ∃ p ∈ N ∗ ∃ p ∈ N ∗ EE p inN^(**)\exists p \in \mathbb{N}^{*}∃p∈N∗ such that lim n → ∞ Δ p − 1 a n = l lim n → ∞   Δ p − 1 a n = l lim_(n rarr oo)Delta^(p-1)a_(n)=l\lim _{n \rightarrow \infty} \Delta^{p-1} a_{n}=llimn→∞Δp−1an=l, where l l lll is finite
b) ( B n ( p , k ) ) B n ( p , k ) (B_(n)(p,k))\left(B_{n}(p, k)\right)(Bn(p,k)) is convergent and lim n → ∞ B n ( p , k ) ≠ 0 lim n → ∞   B n ( p , k ) ≠ 0 lim_(n rarr oo)B_(n)(p,k)!=0\lim _{n \rightarrow \infty} B_{n}(p, k) \neq 0limn→∞Bn(p,k)≠0 with ( B n ( p , k ) ) B n ( p , k ) (B_(n)(p,k))\left(B_{n}(p, k)\right)(Bn(p,k)) defined by (13).
Then ( S n S n S_(n)S_{n}Sn ) converges and the transformation
T n = S n + l − Δ p − 1 a n − k B n − k ( p , k ) , n ∈ N ( p , k ∈ N ) , fixed ) T n = S n + l − Δ p − 1 a n − k B n − k ( p , k ) , n ∈ N ( p , k ∈ N ) ,  fixed  {:T_(n)=S_(n)+(l-Delta^(p-1)a_(n-k))/(B_(n-k)(p,k)),n inNquad(p,k inN)," fixed ")\left.T_{n}=S_{n}+\frac{l-\Delta^{p-1} a_{n-k}}{B_{n-k}(p, k)}, n \in \mathbb{N} \quad(p, k \in \mathbb{N}), \text { fixed }\right)Tn=Sn+l−Δp−1an−kBn−k(p,k),n∈N(p,k∈N), fixed )
accelerates the convergence of ( S n S n S_(n)S_{n}Sn )
Considering x n = Δ p − 1 a n , ∀ n ∈ N x n = Δ p − 1 a n , ∀ n ∈ N x_(n)=Delta^(p-1)a_(n),AA n inNx_{n}=\Delta^{p-1} a_{n}, \forall n \in \mathbb{N}xn=Δp−1an,∀n∈N, the analogy with Theorem 1 is obvious and the results presented there remain true. The usefulness of this formulation results from the following example, in which we use the divergent series ∑ k = 1 n 1 k ∑ k = 1 n   1 k sum_(k=1)^(n)(1)/(k)\sum_{k=1}^{n} \frac{1}{k}∑k=1n1k to prove the convergence of the series ∑ k = 1 ∞ 1 k 2 ∑ k = 1 ∞   1 k 2 sum_(k=1)^(oo)(1)/(k^(2))\sum_{k=1}^{\infty} \frac{1}{k^{2}}∑k=1∞1k2, which is computed.
Numerical example 2
Let S n = ∑ k = 1 n 1 k 2 n ∈ N ∗ S n = ∑ k = 1 n   1 k 2 n ∈ N ∗ S_(n)=sum_(k=1)^(n)(1)/(k^(2))n inN^(**)S_{n}=\sum_{k=1}^{n} \frac{1}{k^{2}} n \in \mathbb{N}^{*}Sn=∑k=1n1k2n∈N∗.
Consider ( a n ) , a n = ∑ k = 1 ∞ 1 k , ∀ n ∈ N ∗ a n , a n = ∑ k = 1 ∞   1 k , ∀ n ∈ N ∗ (a_(n)),a_(n)=sum_(k=1)^(oo)(1)/(k),AA n inN^(**)\left(a_{n}\right), a_{n}=\sum_{k=1}^{\infty} \frac{1}{k}, \forall n \in \mathbb{N}^{*}(an),an=∑k=1∞1k,∀n∈N∗.
This is a divergent sequence, but the sequence ( Δ a n Δ a n Deltaa_(n)\Delta a_{n}Δan ) is convergent, because Δ a n = 1 n + 1 Δ a n = 1 n + 1 Deltaa_(n)=(1)/(n+1)\Delta a_{n}=\frac{1}{n+1}Δan=1n+1 and lim n → ∞ Δ a n = 0 lim n → ∞   Δ a n = 0 lim_(n rarr oo)Deltaa_(n)=0\lim _{n \rightarrow \infty} \Delta a_{n}=0limn→∞Δan=0.
In these conditions, B n ( 2 , 0 ) = Δ 2 a n Δ S n = − n + 1 n + 2 ∀ n ∈ N ∗ B n ( 2 , 0 ) = Δ 2 a n Δ S n = − n + 1 n + 2 ∀ n ∈ N ∗ B_(n)(2,0)=(Delta^(2)a_(n))/(DeltaS_(n))=-(n+1)/(n+2)quad AA n inN^(**)B_{n}(2,0)=\frac{\Delta^{2} a_{n}}{\Delta S_{n}}=-\frac{n+1}{n+2} \quad \forall n \in \mathbb{N}^{*}Bn(2,0)=Δ2anΔSn=−n+1n+2∀n∈N∗ and lim n → ∞ Δ 2 a n Δ S n = − 1 ≠ 0 lim n → ∞   Δ 2 a n Δ S n = − 1 ≠ 0 lim_(n rarr oo)(Delta^(2)a_(n))/(DeltaS_(n))=-1!=0\lim _{n \rightarrow \infty} \frac{\Delta^{2} a_{n}}{\Delta S_{n}}=-1 \neq 0limn→∞Δ2anΔSn=−1≠0.
By Theorem 2 we have that,
T n = S n + l − Δ a n Δ 2 a n Δ S n , n ∈ N ∗ T n = S n + l − Δ a n Δ 2 a n Δ S n , n ∈ N ∗ T_(n)=S_(n)+(l-Deltaa_(n))/((Delta^(2)a_(n))/(DeltaS_(n))),n inN^(**)T_{n}=S_{n}+\frac{l-\Delta a_{n}}{\frac{\Delta^{2} a_{n}}{\Delta S_{n}}}, n \in \mathbb{N}^{*}Tn=Sn+l−ΔanΔ2anΔSn,n∈N∗
accelerates the convergence of ( S n S n S_(n)S_{n}Sn ).
As in Example 1, we construct a transformation
P n = S n + l − ( Δ a n ) b n μ , n ∈ N ∗ P n = S n + l − Δ a n b n μ , n ∈ N ∗ P_(n)=S_(n)+(l-(Deltaa_(n))b_(n))/(mu),quad n inN^(**)P_{n}=S_{n}+\frac{l-\left(\Delta a_{n}\right) b_{n}}{\mu}, \quad n \in \mathbb{N}^{*}Pn=Sn+l−(Δan)bnμ,n∈N∗
which accelerates the convergence of ( S n S n S_(n)S_{n}Sn ) better than
T n = S n + l − Δ a n μ , n ∈ N ∗ T n = S n + l − Δ a n μ , n ∈ N ∗ T_(n)=S_(n)+(l-Deltaa_(n))/(mu),n inN^(**)T_{n}=S_{n}+\frac{l-\Delta a_{n}}{\mu}, n \in \mathbb{N}^{*}Tn=Sn+l−Δanμ,n∈N∗
Table 2
n n nnn S n S n S_(n)S_{n}Sn T n T n T_(n)T_{n}Tn P n P n P_(n)P_{n}Pn
1 1.0000000 1.5000000 1.7500000
2 1.2500000 1.5833333 1.6666667
3 1.3611111 1.6111111 1.6527778
4 1.4236111 1.6236111 1.6486111
5 1.4636111 1.6302778 1.6469444
6 1.4913880 1.6342460 1.6461508
7 1.5117971 1.6367971 1.6457256
8 1.5274221 1.6385332 1.6454776
9 1.5397677 1.6397677 1.6453233
10 1.5497677 1.6406768 1.6452223
11 1.5580322 1.6413655 1.6451534
12 1.5649766 1.6418997 1.6451048
13 1.5708938 1.6423224 1.6450696
14 1.5759958 1.6426625 1.6450435
15 1.5804403 1.6129403 1.6450236
16 1.5843465 1.6431701 1.6450083
17 1.5878067 1.6433623 1.6449963
n S_(n) T_(n) P_(n) 1 1.0000000 1.5000000 1.7500000 2 1.2500000 1.5833333 1.6666667 3 1.3611111 1.6111111 1.6527778 4 1.4236111 1.6236111 1.6486111 5 1.4636111 1.6302778 1.6469444 6 1.4913880 1.6342460 1.6461508 7 1.5117971 1.6367971 1.6457256 8 1.5274221 1.6385332 1.6454776 9 1.5397677 1.6397677 1.6453233 10 1.5497677 1.6406768 1.6452223 11 1.5580322 1.6413655 1.6451534 12 1.5649766 1.6418997 1.6451048 13 1.5708938 1.6423224 1.6450696 14 1.5759958 1.6426625 1.6450435 15 1.5804403 1.6129403 1.6450236 16 1.5843465 1.6431701 1.6450083 17 1.5878067 1.6433623 1.6449963| $n$ | $S_{n}$ | $T_{n}$ | $P_{n}$ | | :--- | :--- | :--- | :--- | | 1 | 1.0000000 | 1.5000000 | 1.7500000 | | 2 | 1.2500000 | 1.5833333 | 1.6666667 | | 3 | 1.3611111 | 1.6111111 | 1.6527778 | | 4 | 1.4236111 | 1.6236111 | 1.6486111 | | 5 | 1.4636111 | 1.6302778 | 1.6469444 | | 6 | 1.4913880 | 1.6342460 | 1.6461508 | | 7 | 1.5117971 | 1.6367971 | 1.6457256 | | 8 | 1.5274221 | 1.6385332 | 1.6454776 | | 9 | 1.5397677 | 1.6397677 | 1.6453233 | | 10 | 1.5497677 | 1.6406768 | 1.6452223 | | 11 | 1.5580322 | 1.6413655 | 1.6451534 | | 12 | 1.5649766 | 1.6418997 | 1.6451048 | | 13 | 1.5708938 | 1.6423224 | 1.6450696 | | 14 | 1.5759958 | 1.6426625 | 1.6450435 | | 15 | 1.5804403 | 1.6129403 | 1.6450236 | | 16 | 1.5843465 | 1.6431701 | 1.6450083 | | 17 | 1.5878067 | 1.6433623 | 1.6449963 |

REFERENCES

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    Heceired 10 Al 1993
1994

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