Prove $f(x)=\sum_{n=0}^{\infty}\frac{1}{1+n^{2}x}$ is continuous on $(0,\,\infty)$ I firstly tried to bound the absolute of difference like this:
$\displaystyle\forall x_{0}\in\left(0\textrm{, }\infty\right)\forall x\in\left(0\textrm{, }\infty\right)\textrm{ and }x\neq x_{0}\,$
$$\left |f(x)-f(x_{0})\right |=\left |\sum_{n=0}^{\infty}\left(\frac{1}{1+n^{2}x}-\frac{1}{1+n^{2}x_{0}}\right)\right |=\left |\sum_{n=1}^{\infty}\frac{n^{2}(x_{0}-x)}{1+n^{2}(x_{0}+x)+n^{4}x_{0}x}\right |\\\leq|x_{0}-x|\sum_{n=1}^{\infty}\frac{n^{2}}{1+n^{4}x_{0}x}\leq\frac{|x_{0}-x|}{x_{0}x}\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}\cdot\frac{|x_{0}-x|}{x_{0}x}$$
However, since $x_{0}x$ can be arbitrarily close to $0$, I cannot bound the last term. 
Can anyone give another approach?
Thanks in advance
 A: Let $f_n(x) = \dfrac{1}{1+n^2x}$. If $x \in [a,b] \subset (0,\infty)$ then $$|f_n(x)| = f_n(x) = \frac{1}{1 + n^2 x} \le \frac{1}{an^2}.$$ The series $$\sum_{n=1}^\infty \frac{1}{an^2}$$ converges, so the Weierstrass $M$-test implies that $$f(x) = \sum_{n=1}^\infty f_n(x)$$ converges uniformly on $[a,b]$. Since each $f_n$ is continuous on $[a,b]$, so is $f$. Since $f$ is continuous on each compact subinterval of $(0,\infty)$ it is continuous on $(0,\infty)$ too.
A: Hint: if $f(x)=\sum_{n=0}^{\infty}\frac{1}{1+n^{2}x}$, $f(\frac{1}{x})=\sum_{n=0}^{\infty}\frac{x}{x+n^{2}}=1+\sum_{n=1}^{\infty}\frac{x}{x+n^{2}}$. $f(x)$ continuous on $(0,\infty)$ if and only if $f(\frac{1}{x})$ continuous on the same interval, and your method applies much more easily to $f(\frac{1}{x})$
A: You're nearly finished. Now note that $x_0$ is constant and positive, and as $|x-x_0|<\delta$ will happen, we see that choosing $\delta<x_0$ small enough ensures that $x_0-\delta<x$ has a positive $x$ bounded away from $0$. Your result is then bounded by

$${\pi^2\over 6}{\delta\over x_0(x_0-\delta)}.$$

Then
$$\delta=\min\left\{{1\over 2}x_0, {6\epsilon x_0^2\over\pi^2}\right\}$$
should suffice.
