Radius of convergence and interval of convergence of $\sum_{n=1}^\infty\frac{(-3)^n}{n \sqrt n}x^n$ $$\sum_{n=1}^\infty\frac{(-3)^n}{n \sqrt n}x^n$$
I tried the ratio test on it but got stuck.
 A: You can use the characterization
$$
R = \sup\{ r\in\mathbb{R}_+ \mid a_n r^n\text{  is bounded} \}
$$
to get that $R=\frac13$.
A: The radius of convergence of the series $\sum a_nx^n$ is given by the formula
$$1/R=\limsup_{n\to\infty}|a_n|^{1/n},$$
which in your case gives
$$1/R=\limsup_{n\to\infty}\left|\frac{(-3)^n}{n \sqrt n} \right|^{1/n}=3,$$
or $R=1/3$.
A: The typical way to compute RoC is
$$\liminf_{n \to \infty} \left [ \frac{n^{3/2}}{3^n}\right ]^{1/n} = \frac{1}{3}$$
A: I'm going to throw this in, since it looks more like what is found in calculus texts:
$$\lim_{n \rightarrow \infty} \ \left|  \frac{a_{n+1}}{a_n}  \right| \ = \ \lim_{n \rightarrow \infty} \ \left|  \frac{(-3)^{n+1} x^{n+1}/ (n+1)^{3/2}}{(-3)^n x^n / n^{3/2}} \  \right|   $$
$$= \ \lim_{n \rightarrow \infty} \ \left| \ \frac{(-3)^{n+1}}{(-3)^n} \ \cdot \ \frac{x^{n+1} }{x^n} \ \cdot \ \left(\frac{ n}{n+1} \right)^{3/2} \ \right| \ = \  \left| \ 3 \ \cdot \ x \ \cdot \ 1 \ \right| \ < \ 1  $$
$$\Rightarrow  \  \vert  x \vert \ < \ \frac{1}{3}  ,  $$
so we need to look at the interval centered on  $ \ x = 0 \ $ with a "radius" of $ \ \frac{1}{3} \ $  .
The "endpoints" must be checked separately.  Using  $ \ x = \frac{1}{3} \ \ \text{and} \ \  x = -\frac{1}{3} \ $ produces series
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{3/2}} \ \text{and} \ \ \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \ . $$
We have a " $p-$series" which is absolutely convergent, so our series converges at both endpoints, making the interval of convergence $ \ \left[  -\frac{1}{3}  ,  \frac{1}{3}  \right] \ . $
