Prove that $ \sum_{n=1}^{\infty} f_n(x)= \sum_{n=1}^{\infty} \frac {1} {n^2 x^2 +1} $ is convergent How do I prove that $ \sum_{n=1}^{\infty} f_n(x)= \sum_{n=1}^{\infty} \frac {1} {n^2 x^2 +1} $ is convergent for every x in real numbers except for $x=0$? 
I tried using the ratio test, but it doesn't seem to be conclusive. 
 A: Assuming you want to know whether $\sum_{n = 1}^\infty\frac{1}{n^2x^2 + 1}$ converges.
Two cases:


*

*$|x| \geq 1$. In this case, we have $$\frac{1}{n^2x^2 + 1} \leq \frac{1}{n^2 + 1} \leq \frac{1}{n^2}$$
and $\sum_{n = 1}^\infty \frac{1}{n^2}$ converges.

*$0<|x| < 1$. In this case, we have
$$
\frac{1}{n^2 x^2 + 1} \leq \frac{1}{n^2x^2 + x^2} = \frac{1}{x^2}\frac{1}{n^2 + 1} \leq \frac{1}{x^2}\cdot\frac{1}{n^2}
$$
and $\sum_{n = 1}^\infty\frac{1}{x^2}\cdot\frac{1}{n^2} = \frac{1}{x^2}\sum_{n = 1}^\infty\frac{1}{n^2}$ converges.


Therefore, the sum $\sum_{n = 1}^\infty\frac{1}{n^2x^2 + 1}$ converges as long as $x \neq 0$.
A: You could use the limit comparison test.
Use $\displaystyle b_n = \frac{1}{n^2}$
Let $\displaystyle a_n = \frac{1}{n^2x^2 + 1}$
$\displaystyle \frac{a_n}{b_n} = \frac{n^2}{n^2x^2 + 1}$
$x$ is a constant $x \ne 0$
$\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} = \frac{1}{x^2}$
Provided $x \ne 0, x^2 > 0, \frac{1}{x^2} > 0$ we find that since $\displaystyle \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2}$ converges to $\displaystyle \frac{\pi^2}{6}$ we find that $\sum a_n$ converges as well. (By Limit Comparison Test)
