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?
 A: 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:
$$\lim \limits_{n \to \infty} \sqrt{\frac{- 2^{-(n+1)} e^{2^{-(n+1)}} \log 2}{-2^{-n} e^{2^{-n}} \log 2}} = \frac{1}{\sqrt{2}} < 1.$$
A: Hint: take the first 4 terms of Maclaurin expansion for $e^{\frac{1}{n}} \sim 1 +\frac{1}{n} +\frac{1}{2n^2} + \frac{1}{6n^3} + O(\frac{1}{n^4})$ and compare to some well-known series or integral. 
A: In general,
for $m \ge 1$,
let
$d_n
=e^{1/n}-\sum_{k=0}^{m-1} \frac1{n^k k!}
$.
Then
$d_n
=\sum_{k=m}^{\infty} \frac1{n^k k!}
=\frac1{m!}\sum_{k=m}^{\infty} \frac{m!}{n^k k!}
$.
Therefore,
$d_n > \frac1{m!n^m}$
and,
if $n \ge 2$,
$d_n 
<\frac1{m!}\sum_{k=m}^{\infty} \frac{1}{n^k }
=\frac1{m!n^m(1-1/n)}
\le\frac{2}{m!n^m}
$.
For $m=3$ (your case),
$\frac1{6n^3}
< d_n
<\frac1{3n^3}
$,
so
$\frac1{\sqrt{6}n^{3/2}}
< d_n^{1/2}
<\frac1{\sqrt{3}n^{3/2}}
$,
and the sum of these converges.
Note that
if you use the cube root,
$\frac1{\sqrt[3]{6}n}
< d_n^{1/3}
<\frac1{\sqrt[3]{3}n}
$,
and the sum of these diverges.
In general,
$\sum d_n^r$
converges for
$r < \frac1{m}$
and diverges for
$r \ge \frac1{m}$.
