I have to check whether this alternating series $\sum\limits_{n=1}^\infty (-1)^n \left(e - \left(1 + \frac{1}{n}\right)^n\right)$ is convergent and absolute convergent. Using the Leibniz criteria I could show convergence. But I got stuck with the absolute convergence ...
I can show $\left|e - \left(1 + \frac{1}{n}\right)^n\right| \leq \frac{e}{n}$ but that would give a divergent upper bound. Anyway my guess is that it is not absolute convergent. But using $\left(1 + \frac{1}{n}\right)^n < \sum\limits_{k=0}^n \frac{1}{k!}$ and then $e - \left(1 + \frac{1}{n}\right)^n \geq \sum\limits_{k=n+1}^\infty \frac{1}{k!}$ leads (I think) to a convergent lower bound.
Can anyone point me in the right direction?
You can use the fact that \begin{align*} \left( {1 + \frac{1}{n}} \right)^n & = \exp \left( {n\log \left( {1 + \frac{1}{n}} \right)} \right) = \exp \left( {n\left( {\frac{1}{n} - \frac{1}{{2n^2 }} + \mathcal{O}\!\left( {\frac{1}{{n^3 }}} \right)} \right)} \right) \\ & = e\exp \left( { - \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) = e - \frac{e}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right). \end{align*}