Show that $\sum_{n=0}^{\infty}xe^{-n^2x}$ converges pointwise on $(0,\infty)$ $\sum\limits_{n=0}^{\infty}xe^{-n^2x} = x\sum\limits_{n=0}^{\infty}e^{-n^2x}$. Apparently the series converges pointwise on $(0,\infty)$ by a limit comparison test, but I cannot see what series I should be comparing this to. What strategy should I use for finding the comparing series?
 A: You may write
$$
\sum_{n=0}^{\infty}e^{-n^2x} = 1+\sum_{n=1}^{\infty}e^{-n^2x}<1+\sum_{n=1}^{\infty}e^{-nx}=1+\frac{1}{e^x-1}<+\infty,\quad x>0.
$$
A: Note that $e^{-n^2x}\le e^{-nx}$, and that for $x\gt 0$ the series $\sum_0^\infty e^{-nx}$ is a geometric series with positive common ratio less than $1$. 
A: $n^2 \geqslant n$, so
$$ e^{-n^2 x} \leqslant e^{-nx}, $$
and
$$ \sum_{n=0}^{\infty} e^{-n^2 x} \leqslant \sum_{n=0}^{\infty} (e^{-x})^n = \frac{1}{1-e^{-x}}. $$
A: By comparison with a Riemann series: it's a series with nonnegative terms, and $\,\mathrm e^{-n^2x}=o\Bigl(\dfrac1{n^2}\Bigr)\,$ if $x>0$.
A: You can compare to $\sum_n e^{-nx}$ which is geometric. 
Another way to do this would be to use the integral test: $\int_{0}^\infty \frac{1}{\sqrt{2 \pi}} e^{-t^2/2} dt = \frac{1}{2}$. By doing a substitution, you can see $\int_0^\infty e^{-t^2 x} dt < \infty$ for $x>0$ fixed, so by integral test you have convergence. 
A: Here's a different approach. Consider the differentiable function 
$$
f_n:[0,\infty)\to[0,\infty),\quad f_n(x)=xe^{-n^2x}.
$$
Since
$$
f_n'(x)=(1-n^2x)e^{-n^2x} \quad \forall x\ge 0,
$$
it follows that $f_n'(x)>0$ for $x\in [0,1/n^2)$, and $f_n'(x)<0$ for $x\in (1/n^2,\infty)$, while $f_n'(1/n^2)=0$. Hence $f_n$ has a maximum at $x=1/n^2$, and therefore 
$$
f_n(x)\le f_n(1/n^2)=\frac{1}{en^2} \quad \forall x\ge0.
$$
Thus, for every $x\ge 0$ we have
$$
\sum_{n=0}^\infty xe^{-n^2x}\le \sum_{n=0}^\infty\frac{1}{en^2}=\frac{\pi^2}{6e}.
$$
