TITLE: For which $x\in \mathbb{R}$ does $\sum_{n=1}^\infty \left(\frac{x^{2n}}{n} - \frac{n^{2x}}{x}\right)$ converge?
QUESTION [6 upvotes]: I have to study for which values of $x \in \mathbb{R}$ the following series converges:

$$\sum_{n=1}^\infty \left(\frac{x^{2n}}{n} - \frac{n^{2x}}{x}\right)$$

I was only able to say that the necessary condition for the convergence of the series, $\left(\frac{x^{2n}}{n} - \frac{n^{2x}}{x}\right) \to 0$, is satisfied iff $-1<x<0$, but then I'm stuck. How would you complete the problem?

REPLY [6 votes]: Since you found the necessary condition $-1 < x < 0$, it follows that
$$\sum_{n=1}^\infty \frac{x^{2n}}{n}$$
converges (absolutely) for all these values of $x$. Thus for $-1 < x < 0$, the series
$$\sum_{n=1}^\infty \biggl(\frac{x^{2n}}{n} - \frac{n^{2x}}{x}\biggr)\tag{1}$$
converges if and only if
$$\sum_{n=1}^\infty \frac{n^{2x}}{x}$$
converges. The constant factor $\frac{1}{x}$ doesn't influence convergence, so you need to see when
$$\sum_{n=1}^\infty n^{2x}$$
converges. It is well-known that that series converges if and only if $2x < -1$, so $x < -\frac{1}{2}$.
Together, we find that the series $(1)$ converges if and only if $-1 < x < -\frac{1}{2}$.
Since
$$\sum_{n=1}^\infty \frac{x^{2n}}{n} = \log \frac{1}{1-x^2}$$
and
$$\sum_{n=1}^\infty n^{2x} = \zeta(-2x),$$
the sum of the series $(1)$ is
$$\sum_{n=1}^\infty \biggl(\frac{x^{2n}}{n} - \frac{n^{2x}}{x}\biggr) = \log \frac{1}{1-x^2} - \frac{\zeta(-2x)}{x}.$$