Prove a series that equals to $\frac{e}{e-1}$ 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?
 A: 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.
