Uniform convergence of $\sum_{n=1}^\infty x^\alpha e^{-nx^2}$ in the set $X=(0;+\infty)$ depending on value of the $\alpha$

Question

I'm trying to explore uniform convergence of series $\sum_{n=1}^\infty x^\alpha e^{-nx^2}$ in the set $X=(0;+\infty)$ depending on value of the $\alpha$.

I tried to use theorem that series are uniformly convergent in the set $X$ if and only if $\lim_{n\to\infty} \sup_{x\in X} |S_n - S| = 0$.

The series converges (normally) to $\dfrac{x^\alpha}{e^{x^2}-1}$, but partial sums are $S_n = \dfrac{x^\alpha}{e^{x^2}-1}\left(1-e^{-nx^2}\right)$, therefore $|S_n-S|=\dfrac{x^\alpha}{e^{nx^2}\left(e^{x^2}-1\right)}$.

To find supremum somehow, I calculated derivative of the expression $\dfrac{x^\alpha}{e^{nx^2}\left(e^{x^2}-1\right)}$ to search for extremum points and got that it is $\dfrac{x^{\alpha-1}\left(\alpha\left(e^{x^2}-1\right)-2nx^2\left(e^{x^2}-1\right)-2x^2e^{x^2}\right)}{e^{nx^2}\left(e^{x^2}-1\right)^2}$.

There are two options, as far as I see, for the derivative to be equal to zero - $x=0$ or $\alpha\left(e^{x^2}-1\right)-2nx^2\left(e^{x^2}-1\right)-2x^2e^{x^2}=0$.

Sadly, I have no idea how to find $x$ from the second equation. Maybe there's some other way to solve this problem?

Answer 1

We have $$\sum x^a e^{-nx^2}=\sum {1\over n^{a/2}}\,(nx^2)^{a/2}\, e^{-nx^2}=\sum {1\over n^{a/2}}\,f(nx^2)$$ where $f(x)=x^{a/2}e^{-x}.$ The maximal value of $f$ is finite for $a>0.$ Hence the series is uniformly convergent if $a>2$ by the Weierstrass M-test.

Concerning $a\le 2,$ I will intentionally avoid using the explicit expression for $S(x)-S_n(x).$ Instead we calculate $$S_{4^{n+1}}(x)-S_{4^n}(x)=\sum_{k=4^n+1}^{4^{n+1}}x^ae^{-kx^2}\ge 3\cdot 4^n\,x^a\,e^{-4^{n+1}x^2}$$ Plug in $x_n=2^{-n}.$ Then $$S_{4^{n+1}}(x_n)-S_{4^n}(x_n)\ge 3\cdot 4^n\, 2^{-an}\, e^{-4}\ge 3\, e^{-4},\quad a\le 2$$ Therefore the difference does not tend to $0$ uniformly when $a\le 2,$ which implies that the subsequence $S(x)-S_{4^n}(x)$ does not converge to $0$ uniformly.

Remark Similar method can be used to study the uniform convergence of the series $$\sum x^a e^{-n^bx^c},\qquad x>0,\ b>0, c>0$$

Answer 2

It's better to use the estimates $$t > 0 \implies \frac{t}{e^t-1} \leq 1 \\ t\approx 0 \implies \frac{t}{e^t-1} \sim 1$$

Using that, you can see $$|S_n-S| \leq x^{\alpha-2}e^{-nx^2}$$ where the estimate is particularly good near $x=0$. Then use whatever method you want (derivatives should work better now) to show $\alpha > 2$ if and only if the convergence is uniform on $(0,\infty)$.