# Does $\frac{n}{\sum\limits_{k=1}^{n}\Big(\frac{k}{k+1}\Big)^n}$ converge?

Does the sequence $$a_n=\displaystyle \frac{n}{\sum\limits_{k=1}^{n}\Big(\frac{k}{k+1}\Big)^n}$$ converge?

Attempt. Since for $k=1,\ldots,n$ $$\frac{1}{2^n}\leq \Big(\frac{k}{k+1}\Big)^n \leq \Big(\frac{n}{n+1}\Big)^n,$$ we get $$\frac{n}{2^n}\leq\frac{1}{a_n}\leq \Big(\frac{n}{n+1}\Big)^n$$ but the sequences $$\frac{n}{2^n},~\Big(\frac{n}{n+1}\Big)^n$$ fail to converge to the same limit. So the squeeze theorem does not work here.

Since the function $$f(x) =\Big(\frac{x}{x+1} \Big)^n$$ increases monotonically, we can use the 'integral comparsion argument' to conlude that $$\int_0^{n} f(x) \, \mathrm{d} x \le \sum_{k=1}^n f(k) \le \int_1^{n+1} f(x) \, \mathrm{d}x.$$ Note that $$\frac{1}{n} \int_0^{n} f(x) \, \mathrm{d} x = \int_0^1 \Big(1+\frac{1}{yn} \Big)^{-n} \mathrm{d} y$$ Since $(1+h/n)^n$ increases monoton in $n$ for all $h \geq 0$, we see that we can apply the dominated convergence theorem to conclude that $$\lim_{n \rightarrow \infty} \int_0^1 \Big(1+\frac{1}{yn} \Big)^{-n} \mathrm{d} y = \int_0^1 \exp(-1/x) \, \mathrm{d} x = \exp(-1) +\mathrm{Ei}(-1).$$ Here we changed coordinates $1/x = y$ and afterwards applied partial integration to get $$\int_0^1 \exp(-1/x) \, \mathrm{d} x = \int_1^\infty \frac{1}{y^2} \exp(-y) \,\mathrm{d} y = e^{-1} + \int_1^\infty \frac{\exp(-x)}{x} \mathrm{d}x = e^{-1} +\mathrm{Ei}(-1).$$ Similarly, we can conclude that $$\frac{1}{n} \int_1^{n+1} f(x) \, \mathrm{d}x = \int_{1/n}^{1+1/n} \Big(1 + \frac{1}{yn}\Big)^{-n} \, \mathrm{d} y \rightarrow \int_0^1 \exp(-1/x) \, \mathrm{d} x.$$ All in all, we see that $$\frac{n}{\sum_{k=1}^n \big( \frac{k}{k+1} \big)^n} \rightarrow \frac{1}{e^{-1} +\mathrm{Ei}(-1)}.$$