Asymptotic Properties and Behavior of Some Nontrivial Sequences

Paul Bracken$^\ast $

August 18, 2020; accepted: November 24, 2020; published online: February 15, 2021.

Convergence properties and asymptotic behavior of several real sequences are investigated analytically. Some remarkable properties of these sequences are established including their limits.

MSC. 33E20, 40A30.

Keywords. sequences, series, monotone, bounded, asymptotic.

$^\ast $Department of Mathematics, University of Texas, Edinburg, TX 78540-2999, paul.bracken@utrgv.edu.

1 Introduction.

The investigation of the general properties and behavior of sequences is a subject that is always of interest and makes use of many tools and ideas from analysis. It is the objective here to study some related sequences none of which is easy to study analytically [ 1 , 2 ] . These sequences are defined on the natural numbers by an analytic formula and lead to many interesting consequences. Asymptotic analysis is very effective, for example as a method of determining limiting behavior [ 2 , 3 ] .

Let us start by defining the two main sequences discussed here at the outset. The first sequence to be studied here is referred to as $( z_n )$ and is defined for all $ n \in \mathbb N$ by the equation

\begin{align} z_n = ( n+1 )^{ \frac{1}{n+1}} - 1 - \tfrac {\log (n)}{n}. \label{eq1} \end{align}

There is another sequence somewhat more untractable than 1 and it is defined for $n \in \mathbb N$ by the formula

\begin{align} x_n = \sum _{k=1}^n \, k^{\frac{1}{k}} - n - \tfrac {1}{2} \log (n)^2. \label{eq2} \end{align}

There is also a version similar to 2 that is involves a definite integral to define it and it is considered last. This sequence is defined for $n \in \mathbb N$ as follows

\begin{align} \beta _n = \int _1^n \, \big( x^{\frac{1}{x}} -1 - \tfrac {\log (x)} {x} \big) \, dx. \label{eq3} \end{align}

These sequences will now be studied in turn. It will be seen that infinite series also play a major role here [ 4 , 5 , 6 ] . There has been interest recently in the study of various kinds of sequences [ 7 , 8 ] . It is also worth mentioning that the asymptotic expansions that appear in theorem 1, for example, can be generated by means of symbolic manipulation.

2 Discrete Sequences.

Theorem 1

The sequence $(z_n)$ consists of terms which are strictly positive and which decrease monotonically from above. The limit of sequence $(z_n)$ as $n \rightarrow \infty $ is zero.

Proof â–¼

The claim can be verified numerically to $n=100$ with $z_{100} \doteq $
$0.00070262816\ldots $ quite easily. Using Maple, the accuracy of numerical calculations can be augmented by setting digits accordingly. For values of $n$ larger than this, $(z_n)$ admits the following asymptotic expansion,

\begin{align} z_n =& \tfrac {1}{n^2} \Big[ ( \tfrac {1}{2} \log (n) -1 ) \log (n) + \tfrac {1}{n} ( \tfrac {1}{6} ( \log (n) - 6) \log (n)^2 + 3 ( 4 \log (n) -3)) \Big] \nonumber \\ & + \tfrac {1}{24 n^4} \Big[ ( \log (n)-12) \, \log (n)^3 + 48 (\log (n) - \tfrac {7}{2} ) \log (n) + 56 \Big]\nonumber \\ & + {\mathcal O} ( \tfrac {\log (n)^5}{n^5} ). \label{eq4} \end{align}

For $n {\gt} 100$, the first term inside the brackets is positive and behaves like $\log (n)^2$. In comparison with the first term, the second and third terms approach zero as $n$ grows. More over, the second term is positive for $\log (n) {\gt}6$, and the third term as well once $\log (n) {\gt} 12$. Combining the numerical work with 4, since $( \log (n) / n)^5 {\lt} 2 \cdot 10^{-7}$ for all $n {\gt}100$, it can be concluded that the sequence $z_n {\gt}0$ for all $n \in \mathbb N$.

To show that $(z_n)$ is a decreasing sequence, define the function

\begin{align} g (x) = (x+1)^{\frac{1}{x+1}} -1 - \tfrac {\log (x)}{x}. \label{eq5} \end{align}

The sequence $( z_n)$ is recovered by putting $x=n$ in 5. Since $g (x)$ has a derivative when $x \in (0, \infty )$, the derivative is found to be

\begin{align} g’ (x) = - ( x+1)^{\frac{1}{x+1}} \big( \tfrac {\log (x+1) -1}{ ( x+1)^2} \big) + \tfrac {\log (x) -1}{x}. \label{eq6} \end{align}

It can be verified that $g' (n) {\lt}0$ for $1 \leq n \leq 100$. Moreover, $g' (n)$ admits the following asymptotic expansion,

\begin{align} g’(n) =& \nonumber - \tfrac {1}{n^3} \big[ ( \log (n) - 3) \log (n) + 3 + \tfrac {1}{2n} ( (\log (n) -7) \log (n)^2 + 16 \log (n) -13 )\nonumber \\ & + \tfrac {1}{6 n^2} ( ( \log (n) -13 ) \log (n)^3 + ( 5 \log (n) -108) \log (n) \big] + {\mathcal O} (\tfrac {\log (n)^5}{n^6}). \label{eq7} \end{align}

The first terms inside the first bracket grows like $\log (n)^2$ whereas the second and third terms become positive too and continue to approach zero as $n$ grows. This implies that the derivative of $g (x)$ is negative, and therefore $z_n$ is a decreasing sequence. From the definition 1, it is clear that in 1 the first two terms and the last approach zero as $n$ gets large and in fact

\begin{align*} \lim _{n \rightarrow \infty } \, z_n = 0. \end{align*}

This proves the claim.

Proof â–¼

The sequence $(x_n)$ is somewhat more challenging and the next theorem begins the study of this sequence.

Theorem 2

The sequence $(x_n)$ defined in 2 has positive values and is strictly increasing for all $n \in \mathbb N$.

Proof â–¼

Consider the difference $x_{n+1} - x_n$. It is calculated from 2 to be exactly

\begin{align} x_{n+1} - x_n = (n +1 )^{\frac{1}{n+1}} -1 - \tfrac {1}{2} \big( \log (n+1)^2 - \log (n)^2 \big) \label{eq8} \end{align}

By the mean value theorem, there exists a $\tau _n \in (n, n+1)$ such that

\begin{align} \log (n+1)^2 - \log (n)^2 = 2 \tfrac {\log (\tau _n)}{\tau _n}. \label{eq9} \end{align}

Since the function $\log (x) /x$ is strictly decreasing for $x {\gt}3$, there are the inequalities

\begin{align} \tfrac {\log (n+1)}{n+1} {\lt} \tfrac {\log (\tau _n)}{\tau _n} {\lt} \tfrac {\log (n)}{n}. \label{eq10} \end{align}

Using 8 and 9 in 7, it is concluded that

\begin{align} x_{n+1} - x_n {\gt} (n+1)^{\frac{1}{n+1}} -1 - \tfrac {\log (n)}{n}. \label{eq11} \end{align}

The right-hand side of 10 is exactly 1, the sequence $(z_n)$. By theorem 1, it is known that $z_n{\gt}0$ for all natural numbers. Applying theorem 1 to 10, it follows that $x_{n+1} - x_n {\gt}0$ or $x_{n+1} {\gt} x_n$. This is stating that sequence $( x_n )$ is strictly increasing. Since $x_1 {\gt}0$ and $(x_n)$ is strictly increasing, it has to be that $x_n {\gt}0$ for all $ n \in \mathbb N$.

Proof â–¼

From 7 it follows that

\begin{align} x_{n+1} {\lt} x_n + (n+1)^{\frac{1}{n+1}} - 1. \label{eq12} \end{align}

The last two terms on the right of 11 approach zero as $n$ gets large. If all $x_n$ can be bounded by a large constant for all $n$ up to $n=N$, then 11 implies that $x_{N+1}$ can be bounded by the same constant. Therefore, $(x_n)$ is strictly increasing and bounded, so it has to converge by the monotone convergence theorem.

In order to obtain the limit, it is necessary to be able to place bounds on certain sums in what follows. To do this it is useful to recall the Euler summation formula.

Theorem 3 Euler Summation Formula

For $m \leq n$,

\begin{equation} \sum _{k=m}^n \, f(k) = \int _m^n \, f(x) \, dx + \tfrac {1}{2} [ f(m) + f(n) ] + \tfrac {1}{12} [ f' (n) - f' (m)] + \rho ( f; m,n), \label{eq13} \end{equation}

with

\[ | \rho (f; m,n) | \leq \tfrac {1}{120} \int _m^n \, | f''' (t) |\, dt. \]

In order to calculate the limit, it is necessary to expand this sequence in the following way. To begin, for $N \geq 1$ we can write

\begin{align} x_N =& \sum _{k=1}^N \, k^{\frac{1}{k}} -N - \tfrac {1}{2} \log (N)^2 = \sum _{k=1}^N \, \sum _{m=0}^{\infty } \, \tfrac {1}{m!} \big( \tfrac {\log k}{k} \big)^m -N - \tfrac {1}{2} \log (N)^2\nonumber \\ =& \sum _{k=1}^N \, \sum _{m=1}^{\infty } \, \tfrac {1}{m!} \big( \tfrac {\log k}{k} \big)^m - \tfrac {1}{2} \log (N)^2\nonumber \\ =& \sum _{k=1}^{N} \, \tfrac {\log k}{k} - \tfrac {1}{2} \log (N) + \sum _{m=2}^{\infty } \, \tfrac {1}{m!} \sum _{k=1}^N \, \big( \tfrac {\log k}{k} \big)^m. \label{eq14} \end{align}

To estimate the $k$-dependent series in 13, the Euler Summation formula is used. For the first series in 13, we have

\begin{align} \sum _{k=1}^N \, \tfrac {\log k}{k} =& \int _1^N \, \tfrac {\log x}{x} \, dx + \tfrac {\log (N)}{2 N} + {\mathcal O} ( \tfrac {\log (N)^2}{N^2} ) = \tfrac {1}{2} \log (N)^2 + \tfrac {\log (N)}{ 2 N}\nonumber \\ & + {\mathcal O} ( \tfrac {\log (N) ^2}{N^2}). \label{eq15} \end{align}

For the last series in 13, the following integral for $m {\gt}1$ is required,

\begin{align} \int _1^N \, \tfrac {(\log x)^m}{x^m} \, dx & = \int _0^{\log (N)} \, u^m e^{- (m+1) u} \, du\nonumber \\ & = \int _0^{\infty } \, u^m e^{- (m-1) u} \, du - \int _{\log (N)}^{\infty } \, u^m e^{- (m-1) u} \, du\nonumber \\ & = \tfrac {m!}{(m-1)^{m-1}} - \int _{\log (N)}^{\infty } \, u^m e^{- (m-1) u} \, du. \label{eq16} \end{align}

For each $m \geq 2$ the remaining integral on the right side of 15 approaches zero as $N \rightarrow \infty $. Consequently, the final sum in $x_N$ takes the following form,

\begin{align} & \sum _{m=2}^{\infty } \, \tfrac {1}{m!} \, \sum _{k=1}^N \, \big( \tfrac {\log k}{k} \big)^m =\nonumber \\ =& \sum _{m=2}^{\infty } \, \tfrac {1}{(m-1)^{m+1} } - \sum _{m=2}^{\infty } \tfrac {1}{m!} \int _{\log (N)}^{\infty } \, u^m e^{- (m-1) u} \, du + {\mathcal O} \big( (\tfrac {\log (N)}{N})^m \big). \label{eq17} \end{align}

Substituting 14 and 16 into 13, we obtain that

\begin{align} x_N = \sum _{n=1}^{\infty } \, \tfrac {1}{n^{n+2}} - \sum _{m=2}^{\infty } \, \tfrac {1}{m!} \int _{\log (N)}^{\infty } \, u^m e^{- (m-1) u} \, du + {\mathcal O} \big(\tfrac {\log (N)}{N} \big). \label{eq18} \end{align}

The result in 17 is presented in the following Theorem.

Theorem 4

The sequence $( x_n )$ for $n \in \mathbb N$ defined by 2 approaches a finite limit which is given by

\begin{align} \lim _{n \rightarrow \infty } \, x_n = \sum _{n=1}^{\infty } \, \tfrac {1}{n^{n+2}} . \label{eq19} \end{align}

3 Integral Version Related to These Sequences.

There is an analogue of the sequence defined in 2 which is worth studying. This sequence employs a definite integral in its definition rather than a sum. To motivate the definition of this sequence, begin by defining the sequence $( I_n )$ in the following form,

\begin{align} I_n = \int _0^n \, ( x^{\frac{1}{x}} -1 ) \, dx. \label{eq20} \end{align}

Writing the function $x^{1/x}$ in the form of an exponential, expanding this exponential and then using some substitutions, equation 19 can be put in the following form,

\begin{align} I_n & = \int _1^n \, ( x^{\frac{1}{x}} -1 ) \, dx = \int _1^n \, \big( \sum _{k=0}^{\infty } \, \tfrac {1}{k!} \big( \tfrac {\log (x)}{x} \big)^k -1 ) \, dx\nonumber \\ & = \int _1^{\infty } \, \sum _{k=1}^{\infty } \, \tfrac {1}{k!} \big( \tfrac {\log (x)}{x} \big)^k \, dx = \int _1^n \, \tfrac {\log (x)}{x} \, dx + \sum _{k=2}^{\infty } \, \tfrac {1}{k!} \int _1^n \, \big( \tfrac {\log (x)}{x} \big)^k \, dx \label{eq21} \\ & = \tfrac {1}{2} \log (n)^2 + \sum _{k=2}^{\infty } \, \tfrac {1}{k!} \int _1^{\log (n)} \, u^k e^{- (k-1) u} \, du\nonumber \\ & = \tfrac {1}{2} \log (n)^2 + \sum _{k=2}^{\infty } \, \tfrac {1}{k!} \tfrac {1}{(k-1)^{k+1}} \int _0^{(k-1) \log (n)} \, t^k e^{-t} \, dt.\nonumber \end{align}

Based on the final equation in 20, since the series on the right has to converge, define the following sequence for $n \in \mathbb N$ already introduced in 3,

\begin{align*} \beta _n = \int _1^n \, \big( x^{\frac{1}{x}} -1 - \tfrac {\log (x)}{x} \big) \, dx. \end{align*}

It is desired to put $\beta _n$ in a form in which its behavior for large $n$ is clear and which implies the limit exists and permits its evaluation. Using the integral form of the gamma function, it is clear that

\begin{align} \beta _n =& \sum _{k=2}^{\infty } \, \tfrac {1}{k!} \tfrac {1}{(k-1)^{k+1}} \Big( \int _0^{\infty } \, t^k e^{-t} \, dt - \int _{(k-1) \log (n)}^{\infty } \, t^k e^{-t} \, dt \Big)\nonumber \\ =& \sum _{k=2}^{\infty } \, \tfrac {1}{(k-1)^{k+1}} - \sum _{k=2}^{\infty } \, \tfrac {1}{k!} \tfrac {1}{(k-1)^{k+1}} \int _{(k-1) \log (n)}^{\infty } \, t^k e^{-t} \, dt \nonumber \\ =& \sum _{k=2}^{\infty } \, \tfrac {1}{(k-1)^{k+1}} - \tfrac {1}{2n} ( \log (n)^2 + 2 \log (n) + 2)\nonumber \\ & - \sum _{k=3}^{\infty } \, \tfrac {1}{k!} \tfrac {1}{(k-1)^{k+1}} \int _{(k-1) \log (n)}^{\infty } \, t^k e^{-t} \, dt. \label{eq22} \end{align}

The second term on the right-hand side approaches zero as $n \rightarrow \infty $. The remaining series on the right side also converges for each large $n$ and in fact approaches zero in the limit as well. It is not hard to give support to this claim by developing the following rough upper bound

\begin{align} & \int _{(k-1) \log (n)}^{\infty } \, t^k e^{-t} e^{- (t-1)} \, dt \leq \nonumber \\ & \leq (( k-1) \log (n) )^k e^{- (k-1) \log (n)} \int _{(k-1) \log (n)}^{\infty } \, e^{-(t-1)} \, dt\nonumber \\ & = \tfrac {(k-1)^k}{n^{k-1}} \, (\log (n))^k \, ( e^{(\log (n))^{-(k-1) }+1} ) = e \, (k-1)^k \tfrac {(\log (n))^k}{(n^{k-1})^2}\nonumber \\ & \leq e \, (k-1)^k \, \tfrac {n^k}{n^{2k-2}}. \label{eq23} \end{align}

Substituting this into the sum in 21, we have the upper bound

\begin{align} \sum _{k=3}^{\infty } \, \tfrac {1}{k!} \tfrac {1}{(k-1)^{k+1}} \int _{(k-1) \log (n)}^{\infty } \, t^k e^{-t} \, dt {\lt} & e \sum _{k=3}^{\infty } \, \tfrac {1}{k! (k-1)} \tfrac {1}{n^{k-2}}\nonumber \\ {\lt}& e \sum _{k=3}^{\infty } \, \tfrac {n^{k-2}}{(k-2)!} {\lt}3 \cdot ( e^{1/n} -1 ). \label{eq24} \end{align}

This implies that the series on the left of 23 is squeezed to zero as $n \rightarrow \infty $. With these results, it is possible to quickly prove the following theorem.

Theorem 5

The sequence $\beta _n$ defined by 3 is increasing and has a finite limit which is given by

\begin{align} \lim _{n \rightarrow \infty } \, \beta _n = \sum _{n=1}^{\infty } \, \tfrac {1}{n^{n+2}}. \label{eq25} \end{align}

Proof â–¼

From the last term in 20, sequence $\beta _n$ is given by

\begin{align} \beta _n = \sum _{k=2}^{\infty } \, \tfrac {1}{k!} \tfrac {1}{(k-1)^{k+1}} \int _0^{(k-1) \log (n)} t^k e^{-t} \, dt. \label{eq26} \end{align}

Hence 25 implies that $\beta _n$ is increasing because the integrand is strictly positive on $(0, \infty )$ and the upper limit on the integral grows like $\log (n)$. Since the right side of the inequality in 23 is finite the remainder term in 21 converges and since it converges to zero as noted already, this implies the limit of sequence $\beta _n$ in 3 exists. Based on result 21, the limit must be given by 24.

Proof â–¼

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