Convergence of $\sum\limits_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) $ I've got in my assignment to show if the following series converges or diverges.
  $$\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) $$


Attempt:
\begin{align*}
\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) &amp;=\sum_{n=1}^\infty \left( 1 + \frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...-1-\frac{1}{n}\right)\\
&amp;=\sum_{n=1}^\infty \left(\frac{1}{2!n^2}+\frac{1}{3!n^3}+...\right)\\
&amp;=\sum_{n=1}^\infty \left(\frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...\right)
\end{align*}

At this point I'm lost. I tried using D'Alambert as follows:

$$\lim \frac{a_{n+1}}{a_n}=\lim\frac{\sqrt[n+1]{e}-1-\frac{1}{n+1}}{\sqrt[n]{e}-1-\frac{1}{n}}$$

which I tried to simplify with basic limit laws (hopefully correctly):

$$\lim\frac{\sqrt[n+1]{e}-1}{\sqrt[n]{e}-1} = 1$$

I don't know where to go from here. Thank you for all your help in advance.
Just use comparison test. Using this Proving Schwarz derivative $\frac{f&#39;&#39;(0)}{2} =\lim\limits_{x\to 0} \frac{\frac{f(x) -f(0)}{x}-f&#39;(0)}{x}$ without Taylor expansion or L&#39;Hopital rule? you have $$\lim_{n\to\infty}{n^2}\left(e^\frac{1}{n} -1 -\frac{1}{n}\right) =\lim_{x\to 0}\frac1{x^2}\left(e^x  -1 -x\right) =\lim\limits_{x\to 0} \frac{\frac{e^x -1}{x}-1}{x}= \frac12$$

Hence, for n large enough we have  $$ \left|n^2\left(e^\frac{1}{n} -1 -\frac{1}{n}\right) -\frac1{2}\right|\le \frac14$$

That is $$ \frac1{4n^2}\le \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) \le \frac3{4n^2}$$
But $$\sum_{n=1}^\infty\frac1{n^2}&lt;\infty$$
 That is 
$$\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right)$$
converges too
Thank you for the comment. By using the comparison test I easily proved the sum converges.

$$\sum_{n=1}^\infty (\frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...) \le \sum_{n=1}^\infty \frac{e}{n^2}$$