uniform convergence for $\sum_{n=1}^\infty t^{\alpha}e^{-nt^2}$

Question

Consider for $\alpha >0$ the series $\sum_{n=1}^\infty t^{\alpha}e^{-nt^2}$

a) Prove that for $\alpha >2$ the series converges uniformly on $[0,\infty)$.
b) Prove that for $0<\alpha\leq 2$, the series doesn't converges uniformly on $[0,\infty)$.

The supremum is at $t=\sqrt{\frac{\alpha}{2n}}$. Therefore you get that $\sum_{n=1}^\infty t^{\alpha}e^{-nt^2}<\sum\frac{\alpha^{\alpha/2}}{2n^{\alpha/2}}e^{-nt^2}=0$ if $\alpha>2$.

How would I prove b) ?

Answer 1

Let $\alpha\geq 0$.

You have perfectly observed the uniform convergence on $[0,+\infty)$ when $\alpha>2$.

Assume that the convergence is uniform on $[0,+\infty)$. Then the sum $$ f(t)=\frac{t^\alpha}{e^{t^2}-1}=\sum_{n\geq 1}t^\alpha e^{-nt^2} $$ is continuous as a uniform limit of continuous function. Note that $$ f(t)\sim \frac{t^\alpha}{t^2}=t^{\alpha-2} $$ at $0$. Since $f$ is continuous at $0$, this forces $\alpha\geq 2$.

So it only remains to consider the case $\alpha=2$. Denote $S_N$ the partial sums of $f$. Then we have $$ f(t)-S_{N-1}(t)=\sum_{n\geq N}t^2e^{-nt^2} $$ In particular, for $t_N=\sqrt{1/N}$, we get $$ \|f-S_{N-1}\|_\infty\geq \sum_{n\geq N}\frac{e^{-n/N}}{N}=\frac{e^{-1}}{N(1-e^{-1/N})}\sim\frac{e^{-1}}{N\frac{1}{N}}=e^{-1}.$$ So the partial sums do not converge uniformly to $f$ on $[0,+\infty)$.

Conclusion: there is uniform convergence on $[0,+\infty)$ if and only if $\alpha>2$.

Answer 2

Let $$f_n(t)=t^\alpha e^{-nt^2}.$$ The supremum of is attained at $t_n=\sqrt{\frac{\alpha}{2n}}$ then since $$||f_n||_\infty=\left(\sqrt{\frac{\alpha}{2}}\right)^\alpha\frac{e^{-\alpha/2}}{n^{\alpha/2}},$$ we have the normal convergence of the series $\sum_n f_n(t)$ which implies the uniform convergence for $\alpha>2$ by comparaison with the Riemann series .

Now, suppose $0<\alpha\leq2$. We have the uniform convergence of the series if $$\lim_{n\to\infty}\sup_{t\in[0,\infty)}|\sum_{k=n+1}^\infty f_k(t)|=0.$$

We have: $$\lim_{n\to\infty}\sup_{t\in[0,\infty)}|\sum_{k=n+1}^\infty f_k(t)|\geq\lim_{n\to\infty}\sum_{k=n+1}^\infty f_k(t_k)=\lim_{n\to\infty}\sum_{k=n+1}^\infty \left(\sqrt{\frac{\alpha}{2}}\right)^\alpha\frac{e^{-\alpha/2}}{k^{\alpha/2}}=+\infty,$$ indeed, the last equality comes from $$\sum_{k=n+1}^\infty\frac{1}{k^{\alpha/2}}\geq\sum_{k=n+1}^\infty\int_{k-1}^k\frac{dx}{x^{\alpha/2}}=\int_n^\infty\frac{dx}{x^{\alpha/2}}=+\infty$$ since $0<\alpha\leq2$.