Convergence of the series $\sum\limits_{n=1}^\infty\frac{\left(1-\frac1n\right)^{n^2}}{3n^2+2}e^n$

Question

Does this series converge? I have tried many methods that I know. $$\sum_{n=1}^\infty\frac{\left(1-\frac1n\right)^{n^2}}{3n^2+2}e^n$$

Edit : after I saw the comments, I tried the ratio method that someone here told me. did I do it right?

Answer 1

First of all, this is a series with positive terms, hence it is enough to show that it is bounded above.

The key is: $$\lim_{n \to \infty} \left( 1- \frac{1}{n} \right)^{n^2} e^n = \frac{1}{\sqrt{e}}$$ in particular, the sequence $\left( 1- \frac{1}{n} \right)^{n^2} e^n$ is convergent, hence it is bounded above by some constant $C > 0$.

Now, $$\sum_{n=1}^{\infty} \left( 1- \frac{1}{n} \right)^{n^2} e^n \frac{1}{3n^2+2} \le \sum_{n=1}^{\infty} \frac{C}{3n^2+2} \le \sum_{n=1}^{\infty} \frac{C}{n^2} < +\infty$$ so that your series is convergent by the comparison test.