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$$ u_{n}=-n+\sum_{k=1}^n e^{\frac{1}{k+n}}$$

$$ u_{n+1}-u_{n}=-1+\exp\left(\frac{1}{2n+2}\right)+\exp\left(\frac{1}{2n+1}\right)-\exp\left(\frac{1}{n+1}\right)=O(1/n^3)$$

So $ \sum u_{n+1}-u_n$ and $u_n$ converge.

$ u_{100000} \approx 0.69 $

The limit seems to be $\ln2$

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Is it $\left[\sum_{k=1}^n e^{\frac{1}{k+n}}\right]-n$ or $\sum_{k=1}^n \left[e^{\frac{1}{k+n}}-n\right]$? – akkkk May 28 '12 at 15:44
Your formula for $u_{n+1}-u_n$ is wrong. – Did May 28 '12 at 15:49
Obviously $u_{n}=-n+\sum_{k=1}^n e^{\frac{1}{k+n}}$ – Chon May 28 '12 at 15:52
I have corrected $u_{n+1}-u_{n}$; the series is convergent... – Chon May 28 '12 at 16:07
Two people have answered, and so far the only up-vote is from me. – Michael Hardy May 28 '12 at 16:11
up vote 5 down vote accepted

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, $\lim\limits_{n\to\infty}u_n=\log2$.

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You can also use $e^x \leq \tfrac{1}{1-x}$ for $x < 1$. This leads to $$ u_n \leq \sum_{k=1}^n \frac{1}{k + n - 1}. $$ – WimC May 28 '12 at 17:19

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.

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