Find the radius of convergence as well as the interval of convergence:

Question

I've done a bunch of these and was successful, but this one is proving to be troubling.

I have no idea how to handle that n*sqrt(n) at the bottom. I ended up with some TROUBLING ITEMS like this lol:

Lim as n approaches infinity = | xnsqrt(n) / (n+1)sqrt(n+1) |

I don't really know how to solve this limit and find out what my R (radius) and I (interval) are. I would isolate the |x| and then just find out what the rest of the problem would be when n is infinity, and end up with something like |x| * 1.

Answer 1

Denote $a_n=\frac{(-5)^n}{n\sqrt n}x^n$. Then $\frac{a_{n+1}}{a_n}=5|x|\frac{n\sqrt n}{(n+1)\sqrt {n+1}} \rightarrow 5|x|$.

Thus the radius of convergence is $(-\frac{1}{5},\frac{1}{5})$.

Note, however, that when $|x|=\frac{1}{5}$, the series is absolutely convergent (recall that $\sum_{n=0}^{\infty}\frac{1}{n \sqrt n}$ converges). So the interval includes the end-points.

Answer 2

Try using the Cauchy-Hadamard Theorem: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem

Answer 3

Using the cauchy hadamard theorem for a power series $\sum_{n=0}^{\infty}a_nx^n$: $$ 1/R=\lim_{n\rightarrow \infty}\sup |a_n|^{1/n} $$ You have for your problem: $$ 1/R=\lim_{n\rightarrow \infty}\sup |\frac{(-5)^n}{n\sqrt{n}}|^{1/n}=5\lim_{n\rightarrow \infty}n^{-1/n^2} $$ $$ =5*1=5\Rightarrow R=1/5 $$ The last limit I evaluated using the log.