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Prove that $$ \lim_{N\to\infty}\sum_{k=0}^\infty \left(1+\frac{k}{N}\right)^{-N}=\frac{e}{e-1} $$

I think $\sum_{k=0}^\infty \left(1+\frac{k}{N}\right)^{-N}$ should be a Riemann sum of a function but could find it. What is the trick in this question?

In addition, the equation holds true when it could interchange the limits, but how to prove it?

Michael Hardy
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Gatsby
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    The result is the one you'd get by interchanging sum and limit, though right now I don't see an evident reason to justify it. –  Aug 08 '16 at 06:48
  • @G.Sassatelli Yes this is what I confused so I tried other ways but gained nothing. – Gatsby Aug 08 '16 at 07:00
  • Hint: $(1+\frac{k}{N})^{-N} = e^{-N}$, when N approach infinity. – Zau Aug 08 '16 at 07:17
  • @ZackNi your argument is correct when interchanging of limits holds true, but can you prove this? – Gatsby Aug 08 '16 at 07:23
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    For every $N$, let $u_N$ denote the function defined on $x>0$ by $$u_N(x)=\left(1+\frac{x}N\right)^{-N}$$ then, note first that $e^{x-x^2}\leqslant1+x\leqslant e^x$ for every $x>0$ hence, for every $k$ and $N$, $$e^{-k}\leqslant u_N(k)\leqslant e^{k^2/N-k}$$ Second, each $u_N$ is decreasing hence, for every $N\geqslant2$ and $n\geqslant1$, $$\sum_{k>n}u_N(k)\leqslant\int_n^\infty u_N(x)dx=\frac{N}{N-1}\left(1+\frac{n}N\right)^{-N+1}\leqslant2u_N(n)\left(1+{}{}{}\frac{n}N\right)$$ Putting these together ... – Did Aug 08 '16 at 07:58
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    ... yields, for every $N\geqslant2$ and $n\geqslant1$, $$\sum_{k=0}^\infty e^{-k}\leqslant\sum_{k=0}^\infty u_N(k)\leqslant e^{n^2/N}\sum_{k=0}^ne^{-k}+2u_N(n)\left(1+\frac{n}N\right)$$ hence $$\frac{1}{1-e^{-1}}\leqslant\sum_{k=0}^{\infty} u_N(k){}{}{}{}\leqslant\frac{e^{n^2/N}}{1-e^{-1}}+2e^{n^2/N}e^{-n}\left(1+\frac{n}{N}\right)$$ If $1\ll n\ll\sqrt{N}$ when $N\to\infty$, one gets $e^{n^2/N}\to1$, $e^{-n}\to0$ and $1+\frac{n}N\to1$ hence $$\lim_{N\to\infty}\sum_{k=0}^\infty u_N(k)=\frac{1}{1-e^{-1}}=\frac{e}{e-1}.$$ – Did Aug 08 '16 at 07:58
  • @Gatsby : I've written a very short answer giving a justification for interchanging the limit and the sum. $\qquad$ – Michael Hardy Aug 08 '16 at 18:22

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Note that $$\left(1+\frac{k}{N}\right)^N = \sum\limits_{j=0}^{N}{{N\choose j}\left(\frac{k}{N}\right)^j} \ge 1 + {N\choose 1}\frac{k}{N} + {N\choose 2}\frac{k^2}{N^2} = 1 + k + \frac{N-1}{2N}k^2 \ge 1 + k + \frac{k^2}{4}$$ for $N\ge 2$. It follows that for any $M\ge 1$ we have $$\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}} - \sum\limits_{k=0}^{M}{\left(1+\frac{k}{N}\right)^{-N}} = \sum\limits_{k=M+1}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}}\le\sum\limits_{k=M+1}^{\infty}{\frac{1}{1+k+k^2/4}}.$$ Note that the RHS goes to zero as $M\rightarrow\infty$. Since $\lim\limits_{N\rightarrow\infty}{\sum\limits_{k=0}^{M}{\left(1+\frac{k}{N}\right)^{-N}}} = \sum\limits_{k=0}^{M}{e^{-k}}$, it follows that \begin{align} &\limsup\limits_{N\rightarrow\infty}{\left(\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}} - \sum\limits_{k=0}^{M}{e^{-k}}\right)} \\ &= \limsup\limits_{N\rightarrow\infty}{\left(\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}} - \sum\limits_{k=0}^{M}{\left(1+\frac{k}{N}\right)^{-N}}\right)}\\ &\le\sum\limits_{k=M+1}^{\infty}{\frac{1}{1+k+k^2/4}} \end{align} i.e. $\limsup\limits_{N\rightarrow\infty}{\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}}}\le \sum\limits_{k=0}^{M}{e^{-k}} + \sum\limits_{k=M+1}^{\infty}{\frac{1}{1+k+k^2/4}}$. Now clearly \begin{align} &\liminf\limits_{N\rightarrow\infty}{\left(\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}} - \sum\limits_{k=0}^{M}{e^{-k}}\right)} \\ &= \liminf\limits_{N\rightarrow\infty}{\left(\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}} - \sum\limits_{k=0}^{M}{\left(1+\frac{k}{N}\right)^{-N}}\right)}\\ &\ge 0 \end{align} i.e. $\liminf\limits_{N\rightarrow\infty}{\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}}}\ge\sum\limits_{k=0}^{M}{e^{-k}}$. Letting $M\rightarrow\infty$ yields $$\limsup\limits_{N\rightarrow\infty}{\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}}}\le\sum\limits_{k=0}^{\infty}{e^{-k}}\le\liminf\limits_{N\rightarrow\infty}{\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}}}\\\implies \lim\limits_{N\rightarrow\infty}{\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}}} = \sum\limits_{k=0}^{\infty}{e^{-k}} = \frac{e}{e-1} $$ as desired.

Joey Zou
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  • Can you tell me how do you reach this step: $\limsup\limits_{N\rightarrow\infty}{\left(\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N})\right)^{-N}} - \sum\limits_{k=0}^{M}{e^{-k}}\right)}\le\sum\limits_{k=M+1}^{\infty}{\frac{1}{1+k+k^2/4}}$ – Gatsby Aug 08 '16 at 10:36
  • Sure. In the line above I establish that $\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}} - \sum\limits_{k=0}^{M}{\left(1+\frac{k}{N}\right)^{-N}}\le\sum\limits_{k=M+1}^{\infty}{\frac{1}{1+k+k^2/4}}$ for all $N$. Taking the $\limsup$ of both sides as $N\rightarrow\infty$, the RHS is independent of $N$ and thus remains constant, while the LHS becomes $\limsup\limits_{N\rightarrow\infty}{\left(\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}} - \sum\limits_{k=0}^{M}{e^{-k}}\right)}$. – Joey Zou Aug 08 '16 at 15:29
  • Answer has been clarified a bit. – Joey Zou Aug 08 '16 at 15:47