real analysis - Convergence of $sum frac{left|sinleft(left(1-frac{1}{n}right)^n-frac{1}{e}right)right|^{alpha}}{e-left(1+frac{1}{n}right)^n}$

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?

Answer

Of course the comparison test is the way to go, one just has to understand how fast $\left(1\pm\frac{1}{n}\right)^n$ converges to $e^{\pm 1}$. For large values of $n$ we have $n\log\left(1\pm\frac{1}{n}\right) = \pm 1 -\frac{1}{2n}+O\left(\frac{1}{n^2}\right)$, hence by exponentiating both sides $\left(1\pm\frac{1}{n}\right)^n \sim e^{\pm 1}\left(1-\frac{1}{2n}\right)+O\left(\frac{1}{n^2}\right)$ and

$$\frac{\left|\sin\left(\left(1-\frac{1}{n}\right)^n-e^{-1}\right)\right|^{\alpha}}{e-\left(1+\frac{1}{n}\right)^n}\sim\frac{\left(\frac{1}{2ne}\right)^{\alpha}}{\frac{e}{2n}}\sim \frac{C_\alpha}{n^{\alpha-1}}$$
so the given series is convergent iff $\alpha>2$.