## Calculus (3rd Edition)

Given $$\sum_{n=1}^{\infty}ne^{-n^2}$$ Since $f(n)=ne^{-n^2},\ \ f'(n) =e^{-n^2}-2e^{-n^2}n^2$, is positive, decreasing and continuous for $n\geq 1$, then by using the integral test \begin{aligned} \int_{1}^{\infty} f(n) dn &=\lim _{t \rightarrow \infty} \int_{1}^{t} ne^{-n^2}d n \\ &=\frac{-1}{2} \left.\lim _{t \rightarrow \infty}e^{-n^2}\right|_{1} ^{t} \\ &= \frac{-1}{2} \lim _{t \rightarrow \infty}e^{-t^2}-1/e\\ &=\frac{1}{2e} \end{aligned} Thus the series converges.