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<h1>Asymptotic Properties and Behavior of Some Nontrivial Sequences</h1>
<p class="authors">
<span class="author">Paul Bracken\(^\ast \)</span>
</p>
<p class="date">August 18, 2020; accepted: November 24, 2020; published online: February 15, 2021.</p>
</div>
<div class="abstract"><p> Convergence properties and asymptotic behavior of several real sequences are investigated analytically. Some remarkable properties of these sequences are established including their limits. </p>
<p><b class="bf">MSC.</b> 33E20, 40A30. </p>
<p><b class="bf">Keywords.</b> sequences, series, monotone, bounded, asymptotic. </p>
</div>
<p>\(^\ast \)Department of Mathematics, University of Texas, Edinburg, TX 78540-2999, <span class="tt">paul.bracken@utrgv.edu</span>. </p>
<h1 id="a0000000002">1 Introduction.</h1>
<p> 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 <span class="cite">
	[
	<a href="#de" >1</a>
	, 
	<a href="#Miller" >2</a>
	]
</span>. 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 <span class="cite">
	[
	<a href="#Miller" >2</a>
	, 
	<a href="#Bracken" >3</a>
	]
</span>. </p>
<p>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 </p>
<div class="displaymath" id="eq1">
  \begin{align}  z_n = ( n+1 )^{ \frac{1}{n+1}} - 1 - \tfrac {\log (n)}{n}. \label{eq1} \end{align}
</div>
<p> There is another sequence somewhat more untractable than <a href="#eq1" class="eqref">1</a> and it is defined for \(n \in \mathbb N\) by the formula </p>
<div class="displaymath" id="eq2">
  \begin{align}  x_n = \sum _{k=1}^n \,  k^{\frac{1}{k}} - n - \tfrac {1}{2} \log (n)^2. \label{eq2} \end{align}
</div>
<p> There is also a version similar to <a href="#eq2" class="eqref">2</a> 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 </p>
<div class="displaymath" id="eq3">
  \begin{align}  \beta _n = \int _1^n \,  \big( x^{\frac{1}{x}} -1 - \tfrac {\log (x)} {x} \big) \,  dx. \label{eq3} \end{align}
</div>
<p> These sequences will now be studied in turn. It will be seen that infinite series also play a major role here <span class="cite">
	[
	<a href="#Knopp" >4</a>
	, 
	<a href="#Hardy" >5</a>
	, 
	<a href="#erd" >6</a>
	]
</span>. There has been interest recently in the study of various kinds of sequences <span class="cite">
	[
	<a href="#duca" >7</a>
	, 
	<a href="#adell" >8</a>
	]
</span>. It is also worth mentioning that the asymptotic expansions that appear in <a href="#th.1">theorem 1</a>, for example, can be generated by means of symbolic manipulation. </p>
<h1 id="a0000000003">2 Discrete Sequences.</h1>
<p><div class="theorem_thmwrapper " id="th.1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1</span>
  </div>
  <div class="theorem_thmcontent">
  <p> 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. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000004">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> The claim can be verified numerically to \(n=100\) with \(z_{100} \doteq \) <br />\(0.00070262816\ldots \) 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, </p>
<div class="displaymath" id="a0000000005">
  \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}
</div>
<p>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 <a href="#eq4" class="eqref">4</a>, 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 \(n \in \mathbb N\). </p>
<p>To show that \((z_n)\) is a decreasing sequence, define the function </p>
<div class="displaymath" id="eq5">
  \begin{align}  g (x) = (x+1)^{\frac{1}{x+1}} -1 - \tfrac {\log (x)}{x}. \label{eq5} \end{align}
</div>
<p> The sequence \(( z_n)\) is recovered by putting \(x=n\) in <a href="#eq5" class="eqref">5</a>. Since \(g (x)\) has a derivative when \(x \in (0, \infty )\), the derivative is found to be </p>
<div class="displaymath" id="eq6">
  \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}
</div>
<p> It can be verified that \(g' (n) {\lt}0\) for \(1 \leq n \leq 100\). Moreover, \(g' (n)\) admits the following asymptotic expansion, </p>
<div class="displaymath" id="a0000000006">
  \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}
</div>
<p>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 <a href="#eq1" class="eqref">1</a>, it is clear that in <a href="#eq1" class="eqref">1</a> the first two terms and the last approach zero as \(n\) gets large and in fact </p>
<div class="displaymath" id="a0000000007">
  \begin{align*}  \lim _{n \rightarrow \infty } \,  z_n = 0. \end{align*}
</div>
<p> This proves the claim. <div class="proof_wrapper" id="a0000000008">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>The sequence \((x_n)\) is somewhat more challenging and the next theorem begins the study of this sequence. </p>
<p><div class="theorem_thmwrapper " id="a0000000009">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p>The sequence \((x_n)\) defined in <a href="#eq2" class="eqref">2</a> has positive values and is strictly increasing for all \(n \in \mathbb N\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000010">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Consider the difference \(x_{n+1} - x_n\). It is calculated from <a href="#eq2" class="eqref">2</a> to be exactly </p>
<div class="displaymath" id="eq8">
  \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}
</div>
<p> By the mean value theorem, there exists a \(\tau _n \in (n, n+1)\) such that </p>
<div class="displaymath" id="eq9">
  \begin{align}  \log (n+1)^2 - \log (n)^2 = 2 \tfrac {\log (\tau _n)}{\tau _n}. \label{eq9} \end{align}
</div>
<p> Since the function \(\log (x) /x\) is strictly decreasing for \(x {\gt}3\), there are the inequalities </p>
<div class="displaymath" id="eq10">
  \begin{align}  \tfrac {\log (n+1)}{n+1} {\lt} \tfrac {\log (\tau _n)}{\tau _n} {\lt} \tfrac {\log (n)}{n}. \label{eq10} \end{align}
</div>
<p> Using <a href="#eq9" class="eqref">8</a> and <a href="#eq10" class="eqref">9</a> in <a href="#eq8" class="eqref">7</a>, it is concluded that </p>
<div class="displaymath" id="eq11">
  \begin{align}  x_{n+1} - x_n {\gt} (n+1)^{\frac{1}{n+1}} -1 - \tfrac {\log (n)}{n}. \label{eq11} \end{align}
</div>
<p> The right-hand side of <a href="#eq11" class="eqref">10</a> is exactly <a href="#eq1" class="eqref">1</a>, the sequence \((z_n)\). By <a href="#th.1">theorem 1</a>, it is known that \(z_n{\gt}0\) for all natural numbers. Applying <a href="#th.1">theorem 1</a> to <a href="#eq11" class="eqref">10</a>, 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 \( n \in \mathbb N\). <div class="proof_wrapper" id="a0000000011">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>From <a href="#eq8" class="eqref">7</a> it follows that </p>
<div class="displaymath" id="eq12">
  \begin{align}  x_{n+1} {\lt} x_n + (n+1)^{\frac{1}{n+1}} - 1. \label{eq12} \end{align}
</div>
<p> The last two terms on the right of <a href="#eq12" class="eqref">11</a> 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 <a href="#eq12" class="eqref">11</a> 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. </p>
<p>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. </p>
<p><div class="theorem_thmwrapper " id="a0000000012">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
    <span class="theorem_thmtitle">Euler Summation Formula</span>
  </div>
  <div class="theorem_thmcontent">
  <p> For \(m \leq n\), </p>
<div class="equation" id="eq13">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">12</span>
</p>
</div>
<p>with </p>
<div class="displaymath" id="a0000000013">
  \[  | \rho (f; m,n) | \leq \tfrac {1}{120} \int _m^n \,  | f''' (t) |\,  dt.  \]
</div>

  </div>
</div> </p>
<p>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 </p>
<div class="displaymath" id="a0000000014">
  \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}
</div>
<p> To estimate the \(k\)-dependent series in <a href="#eq14" class="eqref">13</a>, the Euler Summation formula is used. For the first series in <a href="#eq14" class="eqref">13</a>, we have </p>
<div class="displaymath" id="a0000000015">
  \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}
</div>
<p> For the last series in <a href="#eq14" class="eqref">13</a>, the following integral for \(m {\gt}1\) is required, </p>
<div class="displaymath" id="a0000000016">
  \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}
</div>
<p> For each \(m \geq 2\) the remaining integral on the right side of <a href="#eq16" class="eqref">15</a> approaches zero as \(N \rightarrow \infty \). Consequently, the final sum in \(x_N\) takes the following form, </p>
<div class="displaymath" id="a0000000017">
  \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}
</div>
<p> Substituting <a href="#eq15" class="eqref">14</a> and <a href="#eq17" class="eqref">16</a> into <a href="#eq14" class="eqref">13</a>, we obtain that </p>
<div class="displaymath" id="eq18">
  \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}
</div>
<p> The result in <a href="#eq18" class="eqref">17</a> is presented in the following Theorem. </p>
<p><div class="theorem_thmwrapper " id="a0000000018">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">4</span>
  </div>
  <div class="theorem_thmcontent">
  <p>The sequence \(( x_n )\) for \(n \in \mathbb N\) defined by <a href="#eq2" class="eqref">2</a> approaches a finite limit which is given by </p>
<div class="displaymath" id="eq19">
  \begin{align}  \lim _{n \rightarrow \infty } \,  x_n = \sum _{n=1}^{\infty } \,  \tfrac {1}{n^{n+2}} . \label{eq19} \end{align}
</div>

  </div>
</div> </p>
<h1 id="a0000000019">3 Integral Version Related to These Sequences.</h1>
<p>There is an analogue of the sequence defined in <a href="#eq2" class="eqref">2</a> 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, </p>
<div class="displaymath" id="eq20">
  \begin{align}  I_n = \int _0^n \,  ( x^{\frac{1}{x}} -1 ) \,  dx. \label{eq20} \end{align}
</div>
<p> Writing the function \(x^{1/x}\) in the form of an exponential, expanding this exponential and then using some substitutions, equation <a href="#eq20" class="eqref">19</a> can be put in the following form, </p>
<div class="displaymath" id="a0000000020">
  \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}
</div>
<p> Based on the final equation in <a href="#eq21" class="eqref">20</a>, since the series on the right has to converge, define the following sequence for \(n \in \mathbb N\) already introduced in <a href="#eq3" class="eqref">3</a>, </p>
<div class="displaymath" id="a0000000021">
  \begin{align*}  \beta _n = \int _1^n \,  \big( x^{\frac{1}{x}} -1 - \tfrac {\log (x)}{x} \big) \,  dx. \end{align*}
</div>
<p> 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 </p>
<div class="displaymath" id="a0000000022">
  \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}
</div>
<p> 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 </p>
<div class="displaymath" id="a0000000023">
  \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}
</div>
<p> Substituting this into the sum in <a href="#eq22" class="eqref">21</a>, we have the upper bound </p>
<div class="displaymath" id="a0000000024">
  \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}
</div>
<p> This implies that the series on the left of <a href="#eq24" class="eqref">23</a> is squeezed to zero as \(n \rightarrow \infty \). With these results, it is possible to quickly prove the following theorem. </p>
<p><div class="theorem_thmwrapper " id="a0000000025">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">5</span>
  </div>
  <div class="theorem_thmcontent">
  <p>The sequence \(\beta _n\) defined by <a href="#eq3" class="eqref">3</a> is increasing and has a finite limit which is given by </p>
<div class="displaymath" id="eq25">
  \begin{align}  \lim _{n \rightarrow \infty } \,  \beta _n = \sum _{n=1}^{\infty } \,  \tfrac {1}{n^{n+2}}. \label{eq25} \end{align}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000026">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> From the last term in <a href="#eq21" class="eqref">20</a>, sequence \(\beta _n\) is given by </p>
<div class="displaymath" id="eq26">
  \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}
</div>
<p> Hence <a href="#eq26" class="eqref">25</a> 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 <a href="#eq24" class="eqref">23</a> is finite the remainder term in <a href="#eq22" class="eqref">21</a> converges and since it converges to zero as noted already, this implies the limit of sequence \(\beta _n\) in <a href="#eq3" class="eqref">3</a> exists. Based on result <a href="#eq22" class="eqref">21</a>, the limit must be given by <a href="#eq25" class="eqref">24</a>. <div class="proof_wrapper" id="a0000000027">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="de">1</a></dt>
  <dd><p><i class="sc">N.G. de Bruijn</i>, <i class="it">Asymptotic Methods in Analysis</i>, Dover, Mineola, NY, 1981. </p>
</dd>
  <dt><a name="Miller">2</a></dt>
  <dd><p><i class="sc">P.D. Miller</i>, <i class="it">Asymptotic Methods in Analysis</i>, AMS, Providence, RI, 2006. </p>
</dd>
  <dt><a name="Bracken">3</a></dt>
  <dd><p><a href ="https://doi.org/10.1016/s0723-0869(03)80017-8"> <i class="sc">P. Bracken</i>, <i class="it">Properties of certain sequences related to Stirling’s approximation for the Gamma function</i>, Expo. Math., <b class="bf">21</b> (2003), pp.&#160;171–178. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="Knopp">4</a></dt>
  <dd><p><i class="sc">K. Knopp</i>, <i class="it">Theory and Application of Infinite Series</i>, Dover, Mineola, NY, 1990. </p>
</dd>
  <dt><a name="Hardy">5</a></dt>
  <dd><p><i class="sc">G.H. Hardy</i>, <i class="it">Divergent Series</i>, Clarendon Press, Oxford, UK, 1949. </p>
</dd>
  <dt><a name="erd">6</a></dt>
  <dd><p><i class="sc">A. Erdélyi</i>, <i class="it">Asymptotic Expansions</i>, Dover, Mineola, NY, 1956. </p>
</dd>
  <dt><a name="duca">7</a></dt>
  <dd><p><a href ="https://ictp.acad.ro/jnaat/journal/article/view/1221"> <i class="sc">D. Duca, A. Vernescu</i>, <i class="it">On the convergence rates of the pairs of adjacent sequences</i>, J. Numer. Anal. Approx. Theory, <b class="bf">19</b> (2020), pp.&#160;45–53. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="adell">8</a></dt>
  <dd><p><a href ="https://doi.org/10.1090/mcom/3528 "><i class="sc">J.D. Adell, A. Lakuona</i>, <i class="it">Rational approximation to Euler’s constant at a geometric rate of convergence</i>, Math. Comp., <b class="bf">89</b> (2020), pp.&#160;2553–2561. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
</dl>


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