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

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{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} 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?

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.