Where does the series converges $\sum^\infty_{n=1} (-5)^n \sqrt[5]{\frac{(2n-1)!!}{(2n)!!}}x^{3n}$Is the solution OK?

Question

So guys I want you to tell me is the solution OK. I'm terribly sorry for the not so detailed solution, but the writing in Latex is just too much for me. The calculations aren't that hard so it shouldn't take much time. Just wanted to be sure if my answer is right.

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$\sum^\infty_{n=1} (-5)^n \sqrt[5]{\frac{(2n-1)!!}{(2n)!!}}x^{3n}$, so for $R=\lim_{n \to \inf}|\frac{a_n}{a_{n+1}}|=\frac{1}{5} $, so in the point $-\frac{1}{5}$ and $\frac{1}{5}$ I have to check if it's convergent.For $-\frac{1}{5}$ I get $\lim_{n \to \inf}|\frac{a_n(-\frac{1}{5})}{a_{n+1}(-\frac{1}{5})}|=\frac{1}{25}$, so it converges. For $\frac{1}{5}$ it's basicly the same.So the area must be $[-\frac{1}{5},\frac{1}{5}]$.

Answer 1

There are mistakes in your solution.

• The power of $x$ at the $n^{th}$ term is $x^{3n}$, the radius of convergence is $\frac{1}{\sqrt[3]{5}}$ instead.
• At $x = \pm \frac{1}{\sqrt[3]{5}}$, $\displaystyle\lim_{n\to\infty} \frac{a_n x^{3n}}{a_{n+1} x^{3(n+1)}} = 1$. Ratio test becomes inconclusive and you need to try other criteria.

The situation at $x = \frac{1}{\sqrt[3]{5}}$ is relatively simple. You can verify $\displaystyle \sqrt[5]{\frac{(2n-1)!!}{2n!!}}$ is monotonic decreasing and you can use alternating series test to conclude your series converges there.

The situation at $x = -\frac{1}{\sqrt[3]{5}}$ is more complicated. You now need to show whether

$$\sum_{n=1}^{\infty} \sqrt[5]{\frac{(2n-1)!!}{2n!!}}$$ converges or not. It turns out it didn't. It is easy to check $$\sum_{n=1}^{\infty} \sqrt[5]{\frac{(2n-1)!!}{2n!!}} \ge \sum_{n=1}^{\infty} \frac{(2n-1)!!}{2n!!}\tag{*1}$$ and the series at RHS diverges. Assume the contrary, if the series at RHS converge to some finite number $\lambda$, then notice

$$\sum_{n=1}^\infty \frac{(2n-1)!!}{2n!!} y^n = \frac{1}{\sqrt{1-y}} - 1$$

We can use Abel's theorem for power series to conclude

$$\lim_{y\to 1-} \frac{1}{\sqrt{1-y}} = 1 + \lambda < \infty$$ which is absurd. So the series at RHS and hence the one at LHS of $(*1)$ diverges.

Answer 2

When applying the ratio test, the double-factorial term can be ignored; its ratio from one term to the next approaches $1$. For the remaining piece, the ratio between the $(n+1)$-th term and the $n$-th term is $-5x^3$. The series converges absolutely (diverges) when this is less than (greater than) $1$; i.e., it converges absolutely for $|x|< 5^{-1/3}$ and diverges for $|x|>5^{-1/3}$. The boundary cases, $x=\pm 5^{-1/3}$, require special attention. It looks as if the series diverges for $x=-5^{-1/3}$ (since all terms have the same sign and decrease more slowly than $1/n$) and converges for $x=+5^{-1/3}$ (since the terms are alternating in sign and decreasing in magnitude).