How to solve this limit related to series [closed]

I need just an idea to solve the following limit? I tried everything without success

$$\lim_{N\rightarrow+\infty}\frac{1}{N}\sum_{k=1}^{N-1} \left(\frac{k}{k+1}\right)^N$$

• Way beyond me. Just out of curiosity, is this a made-up problem, an assigned problem, or a [contest or internet] posted problem? That is, do you have reason to believe that the limit can be calculated analytically and then expressed in closed form? Commented May 25, 2021 at 0:43
• @user2661923 5 I added the source Commented May 25, 2021 at 6:18

2 Answers

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.

• Nice, simple (after reading it) and elegant solution ! $\to +1$ for sure. Commented May 25, 2021 at 5:34
• @ClaudeLeibovici Thank you Commented May 25, 2021 at 10:18

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 $$\left( \begin{array}{cc} N & \text{Ei}(-1)+\frac{1}{e}-f_N \\ 10 & 7.68522\times 10^{-4} \\ 100 & 7.66408\times 10^{-6} \\ 1000 & 7.66415\times 10^{-8} \end{array} \right)$$