# Set of $x$ such that $\sum \frac{nx^n}{n^2 + x^{2n}}$ converges

What is the radius of convergence of $$\sum_{n=1}^\infty \frac{nx^n}{n^2 + x^{2n}}?$$ Here is what I have so far. With the ratio test I want to find values of $x$ for which $$\lim_{n\to\infty}x \frac{n+1}{n} \frac{n^2 + x^{2n}}{(n+1)^2 + x^{2(n+1)}} < 1.$$ If $x < 1$, this looks to be true. If $x = 1$, the series is essentially $\sum_{n=1}^\infty \frac{1}{n}$ so diverges. If $x = -1$, $\sum_{n=1}^n \frac{(-1)^n}{n}$ converges by the alternating series test. And for $x > 1$, the series seems to be essentially $$\sum_{n=1}^\infty \frac{1}{x^n}$$ which converges by the ratio test.

• How do you define the radius of convergence for something that is not a power series? Jul 16 '17 at 4:13
• Use root test I think. Jul 16 '17 at 4:16
• I believe the root test is inconclusive. Jul 16 '17 at 4:21
• use the ratio test
– BBot
Jul 16 '17 at 4:27
• Somewhat related: How to resolve this absolute value inequality $|1+x^2|>|x|$? Jul 16 '17 at 8:09

$$\frac{a_{n+1}}{a_{n}} = x\frac{n+1}{n}\frac{n^2 + x^{2n}}{(n+1)^2+x^{2n+2}} =\frac{1}{x}\frac{n+1}{n}\frac{1 + (n/x^n)^2}{1+([n+1]/x^{n+1})^2}$$ If $|x| < 1$, then clearly the first expression goes to $x$ as $n\rightarrow \infty$, and if $|x|>1$, clearly the second expression goes to $x^{-1}$ as $n\rightarrow\infty$. So if $|x| \ne 1$, the series converges absolutely.
What if $x = \pm 1$? Then the series is $\sum_{n=0}^\infty (\pm 1)^n n/(n^2+1)$. If $x = 1$, this series diverges by the integral test against $f(t) = t/(t^2+1)$. If $x = -1$, this series converges by the alternating series test, though only conditionally.
So the series converges for all $x$ except $x = 1$, though only conditionally for $x = -1$.