Regarding a lemma of Tadeusz Ważewski

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Tiberiu Popoviciu
Institutul de Calcul

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T. Popoviciu, À propos d’un lemme de Tadeusz Ważewski, Collection of articles dedicated to the memory of Tadeusz Wazewski, Ann. Polon. Math., 29 (1974) no. 1, pp. 307-308 (in French).

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1974 a -Popoviciu- Ann. Polon. Math. – A propos d’un lemme de Tadeusz Wazewski – scan ICTP

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1974 a -Popoviciu- Ann. Polon. Math. - About a lemma of Tadeusz Wazewski - ICTP scan (1)
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About a lemma of Tadeusz Ważewski by Tiberiu Popoviciu (Cluj)

  1. In most treatises on mathematical analysis we find the well-known theorem, due to Cauchy:
Either ( HAS n HAS n Year)Year}HASn) a given infinite, increasing sequence of positive numbers tending towards + ∞ + ∞ +oo+\infty+∞And ( has n ) has n (year))\left(a_{n}\right)(hasn)any infinite sequence of real numbers. So if we have
(1) lim n → + ∞ has n + 1 − has n HAS n + 1 − HAS n = L (1) lim n → + ∞   has n + 1 − has n HAS n + 1 − HAS n = L {:(1)lim_(n rarr+oo)(a_(n+1)-a_(n))/(A_(n+1)-A_(n))=l:}\begin{equation*} \lim _{n \rightarrow+\infty} \frac{a_{n+1}-a_{n}}{A_{n+1}-A_{n}}=l \tag{1} \end{equation*}(1)limn→+∞hasn+1−hasnHASn+1−HASn=L
we also have
(2) lim n → + ∞ has n HAS n = L (2) lim n → + ∞   has n HAS n = L {:(2)lim_(n rarr+oo)(a_(n))/(A_(n))=l:}\begin{equation*} \lim _{n \rightarrow+\infty} \frac{a_{n}}{A_{n}}=l \tag{2} \end{equation*}(2)limn→+∞hasnHASn=L
The property is also true, whatever the limit L L LLL, finished or not.
Under the assumptions of the theorem, property (2) therefore always results from property (1). On the contrary, property (1) does not always result from property (2). To illustrate this statement, we usually give the simple example: HAS n = n , has n = ( − 1 ) n HAS n = n , has n = ( − 1 ) n A_(n)=n,a_(n)=(-1)^(n)A_{n}=n, a_{n}=(-1)^{n}HASn=n,hasn=(−1)n.
2. An important lemma, established by Tadeusz Ważewski in his remarkable research on l'Hospital's theorem, suggests the following property ( 1 1 ^(1){ }^{1}1):
W. Either ( HAS n ) n = 1 + ∞ HAS n n = 1 + ∞ (A_(n))_(n=1)^(+oo)\left(A_{n}\right)_{n=1}^{+\infty}(HASn)n=1+∞a given infinite, increasing sequence of positive numbers tending towards + ∞ + ∞ +oo+\infty+∞And ( has n ) n = 1 + ∞ has n n = 1 + ∞ (a_(n))_(n=1)^(+oo)\left(a_{n}\right)_{n=1}^{+\infty}(hasn)n=1+∞any infinite sequence of real numbers. Suppose we have property (2). Then we can find a partial sequence ( k n ) n = 1 + ∞ k n n = 1 + ∞ (k_(n))_(n=1)^(+oo)\left(k_{n}\right)_{n=1}^{+\infty}(kn)n=1+∞of the sequence of clues ( n ) n = 1 + ∞ ( n ) n = 1 + ∞ (n)_(n=1)^(+oo)(n)_{n=1}^{+\infty}(n)n=1+∞such that we also have
(3) lim n → + ∞ a k n + 1 − a k n A k n + 1 − A k n = l . (3) lim n → + ∞   a k n + 1 − a k n A k n + 1 − A k n = l . {:(3)lim_(n rarr+oo)(a_(k_(n+1))-a_(k_(n)))/(A_(k_(n+1))-A_(k_(n)))=l.:}\begin{equation*} \lim _{n \rightarrow+\infty} \frac{a_{k_{n+1}}-a_{k_{n}}}{A_{k_{n+1}}-A_{k_{n}}}=l . \tag{3} \end{equation*}(3)limn→+∞haskn+1−hasknHASkn+1−HASkn=L.
We can consider this property as a kind of reciprocal of Oauchy's theorem.
For demonstration of ownership W W WWWwe can follow the proof that Tadeusz Ważewski gave of his lemma.
We have
(4) a k n + 1 − a k n A k n + 1 − A k n = ( a k n + 1 A k n + 1 − a k n A k n ) 1 1 − A k n / A k n + 1 + a k n A k n . (4) a k n + 1 − a k n A k n + 1 − A k n = a k n + 1 A k n + 1 − a k n A k n 1 1 − A k n / A k n + 1 + a k n A k n . {:(4)(a_(k_(n+1))-a_(k_(n)))/(A_(k_(n+1))-A_(k_(n)))=((a_(k_(n+1)))/(A_(k_(n+1)))-(a_(k_(n)))/(A_(k_(n))))(1)/(1-A_(k_(n))//A_(k_(n+1)))+(a_(k_(n)))/(A_(k_(n))).:}\begin{equation*} \frac{a_{k_{n+1}}-a_{k_{n}}}{A_{k_{n+1}}-A_{k_{n}}}=\left(\frac{a_{k_{n+1}}}{A_{k_{n+1}}}-\frac{a_{k_{n}}}{A_{k_{n}}}\right) \frac{1}{1-A_{k_{n}} / A_{k_{n+1}}}+\frac{a_{k_{n}}}{A_{k_{n}}} . \tag{4} \end{equation*}(4)haskn+1−hasknHASkn+1−HASkn=(haskn+1HASkn+1−hasknHASkn)11−HASkn/HASkn+1+hasknHASkn.
But, the sequel ( A n ) A n (A_(n))\left(A_{n}\right)(HASn)tending towards + ∞ + ∞ +oo+\infty+∞, we can choose the continuation ( k n ) k n (k_(n))\left(k_{n}\right)(kn)clues so that we have
(5) A k n A k n + 1 < 1 2 , n = 1 , 2 , … (5) A k n A k n + 1 < 1 2 , n = 1 , 2 , … {:(5)(A_(k_(n)))/(A_(k_(n+1))) < (1)/(2)","quad n=1","2","dots:}\begin{equation*} \frac{A_{k_{n}}}{A_{k_{n+1}}}<\frac{1}{2}, \quad n=1,2, \ldots \tag{5} \end{equation*}(5)HASknHASkn+1<12,n=1,2,…
Let us also note that from (2) it follows that a k n / A k n a k n / A k n a_(k_(n))//A_(k_(n))a_{k_{n}} / A_{k_{n}}haskn/HASkntends towards l l llLpow n → + ∞ n → + ∞ n rarr+oon \rightarrow+\inftyn→+∞.
Property W then follows immediately from (4) when l is finite.
If l l llLis infinite ( = + ∞ = + ∞ =+oo=+\infty=+∞Or − ∞ − ∞ -oo-\infty−∞), we must complete the demonstration. If, for example, l = + ∞ l = + ∞ l=+ool=+\inftyL=+∞, we can first extract a partial sequence ( p n ) n = 1 + ∞ p n n = 1 + ∞ (p_(n))_(n=1)^(+oo)\left(p_{n}\right)_{n=1}^{+\infty}(pn)n=1+∞of clues so that the continuation ( a p n / A p n ) n = 1 + ∞ a p n / A p n n = 1 + ∞ (a_(p_(n))//A_(p_(n)))_(n=1)^(+oo)\left(a_{p_{n}} / A_{p_{n}}\right)_{n=1}^{+\infty}(haspn/HASpn)n=1+∞either increasing (then tending towards + ∞ + ∞ +oo+\infty+∞). We then extract the sequence ( k n k n k_(n)k_{n}kn) of the following ( p n p n p_(n)p_{n}pn), therefore also from the following ( n n nnn) so that we still have (5). We can easily see that this is still possible. Formula (4) shows us that property W is still true for l = + ∞ l = + ∞ l=+ool=+\inftyL=+∞. We demonstrate in the same way the property W W WWWif l = − ∞ l = − ∞ l=-ool=-\inftyL=−∞.
Finally, note that when the limit l l llLis finished, in the property W W WWWwe can choose the partial sequence of indices ( k n k n k_(n)k_{n}kn) independently of the sequence ( a n a n a_(n)a_{n}hasn).
  1. ( 1 ) 1 (^(1))\left({ }^{1}\right)(1)Tadeusz Ważewski, A uniform proof of the generalized l'Hospital theorem, Ann. Soc. Polon. Math. 22 (1949), pp. 161-168.
1974

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