Radius of convergence and interval of convergence of $\sum_{n=1}^\infty\frac{(-3)^n}{n \sqrt n}x^n$

Question

$$\sum_{n=1}^\infty\frac{(-3)^n}{n \sqrt n}x^n$$ I tried the ratio test on it but got stuck.

Answer 1

You can use the characterization $$ R = \sup\{ r\in\mathbb{R}_+ \mid a_n r^n\text{ is bounded} \} $$ to get that $R=\frac13$.

Answer 2

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 $R=1/3$.

Answer 3

The typical way to compute RoC is

$$\liminf_{n \to \infty} \left [ \frac{n^{3/2}}{3^n}\right ]^{1/n} = \frac{1}{3}$$

Answer 4

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] \ . $