# Inequality involving infinite sum

Prove $\sum\limits_{n=1}^\infty ne^{-n^2} \leq \frac{3}{2e}$

Perhaps the mean value theorem for each term may work, but I haven't made any progress. I can't seem to reduce it to a geometric series either.

• Almost there: it is enough to invoke the derivative of a geometric series. Commented Dec 18, 2017 at 21:07

Let $f(x) = x\exp(-x^2)$. Note that $f(x)$ is decreasing for $x \geq 1$. Therefore,
$$\sum_{n\geq 1}n e^{-n^2} = \frac{1}{e}+\sum_{n\geq 2} n e^{-n^2} \leq \frac{1}{e}+\sum_{n\geq 2} n e^{-2n}\leq \frac{6}{5e}\leq\frac{4}{9}$$ since $\sum_{n\geq 2}n x^n = \frac{x^2(2-x)}{(1-x)^2}$ for any $x\in[0,1)$.
• I wonder can we do it like that $\sum\limits_{n=1}^\infty ne^{-n^2}=e^{-1}+\sum\limits_{n=2}^\infty ne^{-n^2}\leq e^{-1}+\displaystyle\int_2^\infty xe^{-x^2}dx=e^{-1}+\frac{1}{2}e^{-4}<e^{-1}+\frac{1}{2}e^{-1}=\frac{3}{2e}.$ Commented Dec 18, 2017 at 21:09
• I don't think it is true that $\sum_{n=2}^{\infty}{n\exp(-n^2)} \leq \int_{2}^{\infty}{x \exp(-x^2)dx}$. Check my solution for a similar idea. Commented Dec 18, 2017 at 21:13