Convergence of $\sum \frac{\left|\sin\left(\left(1-\frac{1}{n}\right)^n-\frac{1}{e}\right)\right|^{\alpha}}{e-\left(1+\frac{1}{n}\right)^n}$ Study the convergence of the following series as $\alpha \in \mathbb{R}$
$$\sum_{n=1}^{\infty}\frac{\left|\sin\left(\left(1-\frac{1}{n}\right)^n-\frac{1}{e}\right)\right|^{\alpha}}{e-\left(1+\frac{1}{n}\right)^n}$$
Maybe this series is quite simple but gives me hard times how to interpret it asymptotically. Could comparison test work?

Hint for the denominator, you can write:
\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}{2 n^2} + o\left(\frac{1}{n^2} \right)\right)\right) \quad (n\rightarrow \infty) \\
&= e^1\left( \exp\left( - \frac{1}{2 n} + o\left(\frac{1}{n} \right)\right) \right)\quad (n\rightarrow \infty) \\
&= e^1\left(1  - \frac{1}{2n} + o\left(\frac{1}{n} \right) \right) \quad (n\rightarrow \infty) \\
\end{align*}
Therefore:
$$ e - \left(1+\frac{1}{n}\right)^n \underset{n\rightarrow\infty}{ \sim \frac{e}{2n} } $$
Using the same method for the numerator, you might be able to discuss the convergence of the series. 
