set of pointwise convergence of series

Question

Find the set of pointwise convergence of series: $$ \sum_{n=0}^{\infty} \frac{3^n}{2n^2+5}x^n $$ I think I need to use Cauchy-Hadamard theorem, but I don't know how to calculate this.

Answer 1

If $\lvert x\rvert\leqslant\frac13$, then$$\left\lvert\frac{3^n}{2n^2+5}x^n\right\rvert\leqslant\frac1{2n^2+5},$$and the series converges (apply the comparison test to this series and to the series $\sum_{n=1}^\infty\frac1{n^2}$).

And if $\lvert x\rvert>\frac13$, then\begin{align}\lim_{n\to\infty}\frac{\left\lvert\frac{3^{n+1}}{2(n+1)n^2+5}x^{n+1}\right\rvert}{\left\lvert\frac{3^n}{2n^2+5}x^n\right\rvert}&=\lim_{n\to\infty}\frac{2n^2+5}{2(n+1)^2+5}\times\frac3{\lvert x\rvert}\\&=\frac3{\lvert x\rvert}\\&>1,\end{align}and so the series diverges then.

Answer 2

Hint: this is a power series around $0$ so it converges to an interval around $0$ and there is a well known formula for the radius of convergence, namely

$$R=\frac{1}{\limsup_{n\to\infty}|a_n|^{1/n}}$$

Compute $R$. Then, you have that the series converges on $(-R,R)$ and that it diverges on $\mathbb{R}-[-R,R]$. Check what happens at points $R$ and $-R$ separately.

answer: $R=1/3$ and the series converges at points $\pm\frac13$ as well.