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) &=\sum_{n=1}^\infty \left( 1 + \frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...-1-\frac{1}{n}\right)\\
&=\sum_{n=1}^\infty \left(\frac{1}{2!n^2}+\frac{1}{3!n^3}+...\right)\\
&=\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.

Thank you for the comment. By using the comparison test I easily proved the sum converges.
$$\sum_{n=1}^\infty (\frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...) \le \sum_{n=1}^\infty \frac{e}{n^2}$$
