Convergence of $\sum_{n=1}^\infty\left(1+\frac{2}{n}\right)^{n^3+n^2+1} \mathrm{e}^{-2n^2}$

Question

Does the series $$ \sum_{n=1}^\infty\left(1+\frac{2}{n}\right)^{n^3+n^2+1} \mathrm{e}^{-2n^2} $$ converge?

The ratio test is inconclusive, so I think I must use the comparison test. But I couldn't find a series to use in this case. Any hint?

Thanks in advance!

Answer 1

Since $$ \log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\mathcal{O}(x^5), \quad \lvert x\rvert<1, $$ we have that $$ \log\left(1+\frac{2}{n}\right)=\frac{2}{n}-\frac{2}{n^2}+\frac{8}{3n^3}-\frac{4}{n^4}+{\mathcal O}(n^{-5}). $$ In particular, using the Taylor expansion theorem, we obtain that the ${\mathcal O}(n^{-5})$ term is of the form $n^{-5}a_n$, whre $a_n$ is a bounded sequence.

Then we have that $$ \log a_n=\log \left(\left(1+\frac{2}{n}\right)^{n^3+n^2+1} \left(\frac{1}{e^2}\right)^{n^2}\right) \\ =(n^3+n^2+1)\left(\frac{2}{n}-\frac{2}{n^2}+\frac{8}{3n^3}-\frac{4}{n^4}+{\mathcal O}(n^{-5})\right)-2n^2\\= \frac{2}{3}+\frac{2}{3n}+{\mathcal O}(n^{-2}) $$ Hence $a_n=\mathrm{e}^{2/3+2/3n+{\mathcal O}(n^{-2})}\to\mathrm{e}^{2/3}$ and thus the series diverges.