Radius of converge of $\sum_{n = 0}^{\infty}\frac{n}{n^{2}\sqrt{n} + 1}x^{n}$

Question

I am supposed to find where the power series:

$$\sum_{n = 0}^{\infty}\frac{n}{n^{2}\sqrt{n} + 1}x^{n}$$ converge absolutely, converge conditionally and where it diverges. My book only shows how to find the radius of convergence by using the ratio test. So I tried it:

$$\lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right| = \lim_{n \to \infty}\left|\frac{(n + 1)(n^{2}\sqrt{n} + 1)}{n((n + 1)^{\frac{5}{2}} + 1)}\right|$$

But I am stuck here, even expanding this more out, I don't know how to find this limit (although I know it is $1$, right?).

Another hint that I got for solving this problem was to use the result I found in the previous task, where I found that $\sum_{n = 0}^{\infty}\frac{n}{n^{2}\sqrt{n} + 1}$ diverges by using the comparison test: $\frac{n}{n^{2}\sqrt{n} + 1} < \frac{1}{n^{3 / 2}}$.

So, how can I find this radius of convergence? How can I use the result that I found?

Answer 1

The limit$$\lim_{n\to\infty}\frac{(n + 1)(n^{2}\sqrt{n} + 1)}{n((n + 1)^{\frac{5}{2}} + 1)}$$is equal to $1$, because it is basically the limit of $\frac{n^{7/2}}{n^{7/2}}$. Therefore, the radius of convergence is $1$.