Convergence of $\sum \frac{\left|\sin\left(\left(1-\frac{1}{n}\right)^n-\frac{1}{e}\right)\right|^{\alpha}}{e-\left(1+\frac{1}{n}\right)^n}$

Question

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 1

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 $\alpha>2$.

Answer 2

By expanding $\ln(1+t)$ and $e^t$ at $t=0$ we get $$\begin{align}\left(1\pm\frac{1}{n}\right)^n&=\exp\left(n\ln\left(1\pm\frac{1}{n}\right)\right)=\exp\left(n\left(\pm\frac{1}{n}-\frac{1}{2n^2}+O(1/n^3)\right)\right)\\ &=e^{\pm1}\left(1-\frac{1}{2n}+O(1/n^2)\right).\end{align}$$ Can you take it from here?

Answer 3

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.