# Checking absolutely convergence

When having an alternating series $$\sum^{\infty}_{n =0} \frac{(-1)^n}{n + x^2}$$ how to check whether one is absolutely convergence or not?

I understand that normally when we have an alternating series with only running index, we can do either ratio test or root test to find out whether the series absolutely converges or not, but this condition how do we separate $n$ and $x$ in order to have a series of the form $$\sum^{\infty}_{n=0} a_n x^n$$ Can someone give me a hint?

• So the issue is whether $\sum \frac{1}{n+x^2}$ converges. It does not, by limit comparison with $\sum \frac{1}{n}$. One could also use the integral test. Remember that here $x^2$ can be treated as a (non-negative) constant. May 10 '16 at 2:25
• Just for your curiosity $\sum^{\infty}_{n =0} \frac{(-1)^n}{n + x^2}=\Phi \left(-1,1,x^2\right)$ where appears the Hurwitz-Lerch transcendent function which is defined everywhere if $x\neq 0$. May 10 '16 at 5:40

For all $n \ge x^2$, we have $2n \ge n+x^2$ so $\frac{1}{2n} \le \frac{1}{n+x^2}.$ Then $$\sum_{n=0}^\infty \left \lvert \frac{(-1)^n}{n+x^2}\right \rvert = \sum^\infty_{n=0} \frac{1}{n+x^2} = \sum_{0\le n < x^2} \frac{1}{n+x^2} + \sum_{n \ge x^2} \frac{1}{n+x^2} \ge \sum_{0\le n < x^2} \frac{1}{n+x^2} + \sum_{n\ge x^2} \frac{1}{2n}.$$ The latter diverges since the harmonic series diverges, so the first sum must diverge as well. Hence it is not absolutely convergent for any $x \in \mathbb R$.
• No. For example $$\sum_{n=0}^\infty \frac{(-1)^{n}}{n^2 + 1}$$ is an alternating series but it converges absolutely. As a general rule, we know an alternating series $$\sum^\infty_{n=0} (-1)^n a_n$$ converges is $a_n \to 0$ as $n \to \infty$; it will converge absolutely if $a_n \to 0$ faster than $\tfrac 1 n \to 0$ as $n \to \infty$. May 10 '16 at 3:38