Determining convergence of $\sum_{n=1}^{+\infty}(e^{\frac{1}{n}}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}$ I have the following infinite series:
$$\sum_{n=1}^{+\infty}(e^{\frac{1}{n}}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}$$
I want to examine its convergence. First thing that came to my mind was "unfolding" $e^{\frac{1}{n}}$ and see what will happen:
$$\sum_{n=1}^{+\infty}(e^{\frac{1}{n}}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}=\sum_{n=1}^{+\infty}(\sum_{k=0}^{+\infty}\frac{1}{n^kk!}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}=\sum_{n=1}^{+\infty}(\sum_{k=3}^{+\infty}\frac{1}{n^kk!})^{\frac{1}{2}}$$
Now it kinda "looks" convergent but I'm clueless about how to prove it. Any hints?

We have
$$\sum_{n=1}^{\infty} \sqrt{\sum_{k=3}^{\infty} \frac{1}{n^k k!}}$$
We can use the Cauchy condensation test to study the convergence of
$$\sum_{n=1}^{\infty} 2^n \sqrt{\sum_{k=3}^{\infty} \frac{1}{2^{nk} k!}} = \sum_{n=1}^{\infty} 2^n \sqrt{\frac{1}{2^{2n}} \sum_{k=3}^{\infty} \frac{1}{2^{n(k-2)} k!}}$$
$$\sum_{n=1}^{\infty} \sqrt{\sum_{k=3}^{\infty} \frac{1}{2^{n(k-2)} k!}} \leq \sum_{n=1}^{\infty} \sqrt{\sum_{k=1}^{\infty} \frac{1}{2^{nk} k!}} = \sum_{n=1}^{\infty} \sqrt{e^{2^{-n}} - 1}$$ 
The final series converges by the ratio test:
$$\lim \limits_{n \to \infty} \sqrt{ \frac{e^{2^{-(n+1)}} - 1}{e^{2^{-n}} - 1} }$$
This limit goes to $0/0$, so use L'Hopital's on the inside:
$$\lim \limits_{n \to \infty} \sqrt{\frac{- 2^{-(n+1)} e^{2^{-(n+1)}} \log 2}{-2^{-n} e^{2^{-n}} \log 2}} = \frac{1}{\sqrt{2}} < 1.$$
