Examining convergence of $\sum_{n=1}^{\infty}\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}$ with mean value theorem $$\sum_{n=1}^{\infty}\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}$$
I would like to examine covergence of this series using mean value theorem.  
I would like to check my solution and optionally alternative solution :)  
Let $c_n \in (n, n+1)$.
$f(n)=e^{\frac{1}{n}}$.
$$\frac{e^{\frac1n}-e^{\frac1{n+1}}}{-1}=\left(e^{\frac{1}{c_n}}\right)'=-\frac{1}{c_n^2}e^{\frac1{c_n}}$$
 $$\sqrt{e^{\frac1n}-e^{\frac{1}{n+1}}}=\sqrt{\frac{1}{c_n^2}e^{\frac{1}{c_n}}}=\frac{1}{c_n}\sqrt{e^{\frac{1}{c_n}}}$$
$$\sum_{n=1}^{\infty}\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}\le \sum_{n=1}^{\infty}\frac{1}{n^2}\sqrt{{e^{\frac{1}{n}}}}$$
$$\sum_{n=1}^{\infty}\frac{1}{n^2}\sqrt{{e^{\frac{1}{n}}}}\le \sum_{n=1}^{\infty}\frac{1}{n^2}{e^{\frac{1}{n}}}$$    Since, is $\sum_{n=1}^{\infty}\frac{1}{n^2}{e^{\frac{1}{n}}}$ covergent absolutely so our seires is also absolutely covergent. 
 A: Doing basically the same as Olivier Oloa in his answer with Taylor series of higher order $$e^{\frac{1}{n }}=1+\frac{1}{n}+\frac{1}{2 n^2}+\frac{1}{6 n^3}+\frac{1}{24
   n^4}+O\left(\frac{1}{n^5}\right)$$ $$e^{\frac{1}{n+1 }}=1+\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{6 n^3}+\frac{1}{24
   n^4}+O\left(\frac{1}{n^5}\right)$$ $$e^{\frac{1}{n }}-e^{\frac{1}{n+1 }}=\left(\frac{1}{n}\right)^2+O\left(\frac{1}{n^5}\right)$$ $$\sqrt{e^{\frac{1}{n }}-e^{\frac{1}{n+1 }}}=\frac{1}{n}+O\left(\frac{1}{n^4}\right)$$ $$\frac 1 n \sqrt{e^{\frac{1}{n }}-e^{\frac{1}{n+1 }}}=\frac{1}{n^2}+O\left(\frac{1}{n^5}\right)$$ Using one extra order, we should have get $$\frac 1 n \sqrt{e^{\frac{1}{n }}-e^{\frac{1}{n+1 }}}=\frac{1}{n^2}+\frac{1}{12
   n^5}+O\left(\frac{1}{n^6}\right)$$
A: It is OK. Alternatively, you may write, by the Taylor series expansion, as $n \to \infty$, 
$$
\begin{align}
\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}&=\frac{1}{n}\sqrt{\left(1+O\left(\frac{1}{n}\right)\right)-\left(1+O\left(\frac{1}{n+1}\right)\right)}
\\\\&=\frac{1}{n}\sqrt{O\left(\frac{1}{n}\right)}
\\\\&=O\left(\frac{1}{n^{3/2}}\right)
\end{align}
$$ giving the convergence of the initial series. We have used the fact that, as $n \to \infty$,
$$
O\left(\frac{1}{n+1}\right)=O\left(\frac1n\cdot\frac{1}{1+1/n}\right)=O\left(\frac1n\cdot(1-1/n)\right)=O\left(\frac{1}n\right).
$$
