How to solve this limit related to series I need just an idea to solve the following limit? I tried everything without success
$\lim_{N\rightarrow+\infty}\frac{1}{N}\sum_{k=1}^{N-1} \left(\frac{k}{k+1}\right)^N$
Source https://artofproblemsolving.com/community/c7h495716p2783297
 A: Start with the following estimation:
$$(1+\frac{1}{n})^n \le e \le (1+\frac{1}{n})^{n+1} \quad \forall n \ge 1$$
So
$$\sum_{n=1}^{N-1} e^{-N/(n+1)} \ge \sum_{n=1}^{N-1} \dfrac{1}{ \left( 1+\frac{1}{n}\right)^N } \ge \sum_{n=1}^{N-1} e^{-N/n}$$
Thus
$$\lim_{N \rightarrow +\infty} \frac{1}{N}\sum_{n=1}^{N-1} \dfrac{n^N}{ \left( n+1\right)^N } = \lim_{N \rightarrow +\infty} \frac{1}{N}\sum_{n=1}^{N-1} e^{-N/n} $$
Besides
$$\lim_{N \rightarrow +\infty} \frac{1}{N}\sum_{n=1}^{N-1} e^{-N/n}=\int_{0}^1 e^{-1/x}dx $$
So the limit value is $\int_{0}^1 e^{-1/x}dx $.
Note: This value has no concise form.
A: Starting from @Paresseux Nguyen's elegant solution
$$\lim_{N \rightarrow +\infty} \frac{1}{N}\sum_{n=1}^{N-1} e^{-N/n}=\int_{0}^1 e^{-1/x}dx=\text{Ei}(-1)+\frac{1}{e}\sim 0.1484955068$$ where appears the exponential integral function.
If
$$f_N=\frac{1}{N}\sum_{k=1}^{N-1} \left(\frac{k}{k+1}\right)^N$$ Since $f_2=\frac 18=0.125$, I was expecting a rather fast convergence; this is not the case
$$\left(
\begin{array}{cc}
N & \text{Ei}(-1)+\frac{1}{e}-f_N \\
 10 & 7.68522\times 10^{-4} \\
 100 & 7.66408\times 10^{-6} \\
 1000 & 7.66415\times 10^{-8}
\end{array}
\right)$$
