Convergence or divergence of $ u_{n}=\left(\sum\limits_{k=1}^n e^{\frac{1}{k+n}}\right)-n$ $$ u_{n}=-n+\sum_{k=1}^n e^{\frac{1}{k+n}}$$
$$ u_{n+1}-u_{n}=-1+\exp\left(\frac{1}{2n+2}\right)+\exp\left(\frac{1}{2n+1}\right)-\exp\left(\frac{1}{n+1}\right)=O(1/n^3)$$
So $ \sum u_{n+1}-u_n$ and $u_n$ converge.
$ u_{100000} \approx 0.69 $
The limit seems to be $\ln2$
 A: For every $x\leqslant1$, $1+x\leqslant\mathrm e^x\leqslant1+x+x^2$ hence $u_n=v_n+w_n$ with
$$
v_n=\sum_{k=1}^n\frac1{k+n}=\frac1n\sum_{k=1}^n\frac1{\frac{k}n+1},\qquad 0\leqslant w_n\leqslant\sum_{k=1}^n\frac1{(k+n)^2}\leqslant\frac1n.
$$
In particular, $w_n\to0$. On the other hand, $v_n$ is the $n$th Riemann sum of the function $x\mapsto\frac1{1+x}$ on the interval $[0,1]$ hence $v_n\to\int\limits_0^1\frac{\mathrm dx}{1+x}=\log2$. Finally, $\lim\limits_{n\to\infty}u_n=\log2$.
A: Everything boils down to the n times use of the following elementary limit, namely:
$$\lim_{x\to0}\frac{e^{x}-1}{x}=1$$
Consequently, by expanding the sum we have that:
$$\lim_{n\to\infty}\frac{e^\frac{1}{n+1}-1}{\frac{1}{n+1}} \frac{1}{n+1} + \lim_{n\to\infty}\frac{e^\frac{1}{n+2}-1}{\frac{1}{n+2}} \frac{1}{n+2}+ \cdots = \lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} = $$
$$\lim_{n\to\infty}{\gamma}+\ln{2n}-{\gamma}-\ln{n}= \ln{2}.$$
Also notice that i've just applied another well-known limit:  $$\lim_{n\to\infty} 1+\frac1{2}+\cdots+\frac{1}{n}-\ln{n}={\gamma}$$ $\tag{$\gamma$ is Euler-Mascheroni constant}$
The proof is complete.
