Does $\sum_{n=1}^{\infty} (-1)^n \left[e-\left(1+\frac{1}{n}\right)^n \right]$ converge absolutely?

Question

Check whether the series $\sum_{n=1}^{\infty} (-1)^n \left[e-\left(1+\frac{1}{n}\right)^n \right]$ converge absolutely?

What I attempted:-

If $a_n=(-1)^n \left[e-\left(1+\frac{1}{n}\right)^n \right]$, then by Leibnitz test $\sum_{n=1}^{\infty}a_n$ converges since $\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n=e$ .

Now, $|a_n|=\left[e-\left(1+\frac{1}{n}\right)^n \right]$. Thus, \begin{equation} \begin{aligned} \frac{|a_{n+1}|}{|a_n|}&=\frac{e-\left(1+\frac{1}{n+1}\right)^{n+1}}{e-\left(1+\frac{1}{n}\right)^n} \end{aligned} \end{equation} For large $n$, this ratio takes the form of $\frac{0}{0}$. Moreover, it is not difficult to see that $\left(|a_n|\right)_{n=1}^{\infty}$ form a diminishing sequence. Hence, intuition may lead to answer that it converges absolutely, which, I think is not true. I am not getting any way to proceed.
A hint will be highly appreciated. How could we suspect it to be divergent or convergent?

Answer 1

It's well-known that the sequence $\left(1+\frac1n\right)^n$ increases with limit $e$. So, you question is "Does $\sum_{n=1}^\infty b_n$ converge where $b_n=e-\left(1+\frac1n\right)^n$?".

Now $$\ln\left(1+\frac 1n\right)^n=n\ln\left(1+\frac1n\right)=n\left( \frac1n-\frac1{2n^2}+O(n^{-3})\right) =1-\frac1{2n}+O(n^{-2}).$$ Exponentiating, $$\left(1+\frac 1n\right)^n=\exp\left(1-\frac1{2n}+O(n^{-2})\right) =e\left(1-\frac1{2n}+O(n^{-2})\right).$$ Then $$b_n=\frac{e}{2n}+O(n^{-2}).$$ Therefore $\sum b_n$ diverges, although $\sum (-1)^nb_n$ converges.