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.
Thanks in advance.

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
$$\frac{n}{\sum_{k=1}^n \big( \frac{k}{k+1} \big)^n} \rightarrow \frac{1}{e^{-1} +\mathrm{Ei}(-1)}.$$
