How to compute $\lim_{n\to \infty}n\left( \frac{e}{e-1}-\sum_{k=1}^n\left(\frac{k}{n}\right)^n\right)$?

Question

• By reading What is $\lim_{n\to \infty}\sum_{k=1}^n \left(\frac{k}{n}\right)^n$? I have already known that $$ \lim_{n\to \infty} \sum_{k=1}^n\left(1-\frac{k}{n}\right)^n=\frac{e}{e-1} $$
• But then a further question comes, what's $\displaystyle{% \lim_{n\to \infty}\left\{n\left[{\mathrm{e} \over \mathrm{e} - 1} - \sum_{k = 1}^{n}\left(k \over n\right)^{n}\right]\right\}}$ ?.

Can anyone give some suggestions ?.

Answer 1

Let $a_n$ be the sequence at hand, we can rewrite it into following form $$ a_n \stackrel{def}{=} n\left[\frac{e}{e-1} - \sum_{k=1}^n\frac{k^n}{n^n}\right] = n\left[\frac{1}{1-e^{-1}} - \sum_{k=0}^{n-1}\left(1-\frac{k}{n}\right)^n\right] = n\left[\frac{e^{-n}}{1-e^{-1}} + \sum_{k=0}^{n-1} b_{n,k} e^{-k}\right] $$ where $$b_{n,k} \stackrel{def}{=} 1 - \left(1-\frac{k}{n}\right)^n e^k $$

For fixed $k$, it is easy to show $b_{n,k} \in [0,1]$ whenever $n > k$. In fact, this sequence monotonically decreases to $0$ as $n \to \infty$.

Take $\alpha \in (0,\frac13)$ and $K = \lfloor n^\alpha \rfloor$. Since $b_{n,k} \in [0,1]$, we have

$$\left|\sum_{k=K+1}^{n-1} b_{n,k} e^{-k}\right| \le \sum_{k=K+1}^{n-1} e^{-k} < \sum_{k=K+1}^\infty e^{-k} = \frac{e^{-K}}{e - 1} $$ From this, we find

$$a_n = \sum_{k=0}^K n b_{n,k} e^{-k} + O( n e^{-n^\alpha} )$$

For $k \le K$, we have

$$\begin{align} n b_{n,k} &= n(1 - e^{k+n\log(1 - k/n)}) = n\left(1 - e^{-\left(\frac{k^2}{2n} + \frac{k^3}{3n^2} + \cdots\right)}\right)\\ &= n\left(1 - e^{-\frac{k^2}{2n} + O(n^{3\alpha-2})}\right) = \frac{k^2}{2} + O(n^{3\alpha-1}) \end{align}\tag{*1} $$ Since as a sequence in $K$, $\sum_{k=0}^K e^{-k}$ is bounded and the error term in RHS of $(*1)$ can be bounded in a manner independent of $k$, we have

$$\sum_{k=0}^K n b_{n,k} e^{-k} = \sum_{k=0}^K \frac{k^2}{2} e^{-k} + O(n^{3\alpha-1}) \quad\implies\quad a_n = \sum_{k=1}^{\lfloor n^\alpha \rfloor} \frac{k^2}{2} e^{-k} + O(n^{3\alpha-1}) $$ As a result, $$\lim_{n\to\infty} a_n = \sum_{k=1}^\infty \frac{k^2}{2} e^{-k} = \frac{e^{-1}(1+e^{-1})}{2(1-e^{-1})^3} = \frac{e(e+1)}{2(e-1)^3} \approx 0.9961473835624938 $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\begin{equation} \lim_{n\to \infty}\braces{n\bracks{{\expo{} \over \expo{} - 1} - \sum_{k = 1}^{n}\pars{k \over n}^{n}}}:\ ? \label{1}\tag{1} \end{equation}

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The main contribution to the sum $\ds{\sum_{k = 1}^{n}k^{n}}$ ocurrs for values of $\ds{k \approx n}$. In order to derive an asymptotic approximation as $\ds{n \to \infty}$ we can use the $\color{#f66}{Laplace\ Methods\ for\ Sums}$ as explained in, for example, ( $\color{#44f}{Analytic Combinatorics}$ by Philippe Flajolet and Robert Sedgewick, page $761$, Cambridge University Press 2009 ). \begin{align} \left.{1 \over n^{n}}\sum_{k = 1}^{n}k^{n}\,\right\vert_{\ n\ >\ 0} & = {1 \over n^{n}}\sum_{k = 0}^{n}\pars{n - k}^{n} = \sum_{k = 0}^{n}\pars{1 - {k \over n}}^{n} = \sum_{k = 0}^{n} \exp\pars{n\ln\pars{1 - {k \over n}}} \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, \sum_{k = 0}^{\infty} \expo{-k}\pars{1 - {k^{2} \over 2n}} + \,\mrm{O}\pars{1 \over n^{2}} \\[5mm] & = {\expo{} \over \expo{} - 1} - \color{#f00}{\expo{}\pars{\expo{} + 1} \over 2\pars{\expo{} - 1}^{3}} \,{1 \over n} + \,\mrm{O}\pars{1 \over n^{2}} \end{align} such that the requested limit, in \eqref{1}, becomes $$\bbox[15px,#ffe,border:1px groove navy]{\ds{% \lim_{n\to \infty}\braces{n\bracks{{\expo{} \over \expo{} - 1} - \sum_{k = 1}^{n}\pars{k \over n}^{n}}} = {\expo{}\pars{\expo{} + 1} \over 2\pars{\expo{} - 1}^{3}} \approx 0.99614738356249369646\ldots}} $$