I have the following series:
$$\sum\limits_{n=1}^\infty\left(e^\frac{1}{n} - e^\frac{1}{n+2}\right)$$
I know that this series converges to $e + \sqrt{e} - 2$ (I found its sum using $S = \lim\limits_{n \to \infty} S_n$, where $S_n$ is a partial sum of $n$ elements).
But is there any way to prove that the series converges without finding the actual sum?
$f(x) = e^x$ satisfies $f'(x) = e^x < e$ for $0 < x < 1$, so that by the mean-value theorem $$ 0 < e^\frac{1}{n} - e^\frac{1}{n+2} < e \left( \frac 1n - \frac{1}{n+2}\right) = \frac{2e}{n(n+2)} < \frac{2e}{n^2} \, . $$ This shows that the series converges by comparison with $\sum 1/n^2$.
We have that
$$ \left(1+\frac1n\right)^n\le e \le \left(1+\frac1n\right)^{n+1}$$
and then
$$e^\frac{1}{n} - e^\frac{1}{n+2}\le \left[\left(1+\frac1{n-1}\right)^n\right]^\frac1n-\left[\left(1+\frac1{n+2}\right)^{n+2}\right]^\frac1{n+2} =\frac1{n-1}-\frac1{n+2}=\frac3{(n-1)(n+2)}$$
therefore the series converges.
As noticed by MartinR in the comments, the same estimate follows also from the well-known inequality $1+x \le e^x \le 1/(1-x)$, indeed
$$e^\frac{1}{n} - e^\frac{1}{n+2}\le \frac1{1-\frac1n}-1-\frac{1}{n+2}=\frac3{(n-1)(n+2)}$$
At infinity, using Taylor expansion, you have: $$e^{\frac1n}-e^{\frac1{n+2}}\sim 1+\frac1n-1-\frac1{n+2}=\frac{2}{n(n+2)}\sim\frac2{n^2} $$ Hence the series is convergent.
Another way, using that by standard limit $\frac{e^x-1}x\to 1$ we have
$$e^\frac{1}{n} - e^\frac{1}{n+2}=e^\frac{1}{n+2}\;\frac{e^\frac{2}{n(n+2)}-1}{\frac{2}{n(n+2)}}\;\frac{2}{n(n+2)}\sim \frac{2}{n(n+2)}$$
