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?

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.
