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Consider $f(x)=\sum_{n=1}^{\infty} \frac{1}{1+n^2 x}.$ For what values of x does it converge uniformly?

In general is there any special criterion, except Weierstrass M-test, to find out whether a Series of functions is uniformly convergent or not?

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    The series is UC for $x\ge x_0>0$. – 2017-01-03
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    @ guru did you say that pointwise convergence implies uniform convergence? – 2017-01-03
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    @guru You should consider defining "Normal Convergence" for the OP, else the word "normal convergence" might be interpreted to mean "convergence." – 2017-01-03
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    @manhattan Normal here means the sup norm. – 2017-01-03
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    I can see pointwise convergence but I do not know how to prove uniform convergence. In general in series I have problem with proving uniform convergence. Do you have any idea how can I learn this topic? – 2017-01-03

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Suppose $x\geq 0$ to avoid values like $\frac{-1}{4}, \frac{-1}{9},...$.

For $a>0$, the series converges uniformly at $[a,+\infty)$ by M-test. it doesn't at $[0,+\infty)$ cause if the concergence is uniform at $[0,+\infty)$, the sum function will be continuous at $x=0$ cause the functions $x\mapsto \frac{1}{1+n^2x}$ are all continuous at $x=0$. but for $x=0$, the series diverges.

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The series converges point wise for $x>0$ as well as for $x<0$ as long as $-\frac1x$ is not a perfect square (so in particular for $x<-1$). Treating the negative part is a bit tricky, but on $(0,\infty)$, we see that $0\le f(x)\le\frac {\pi^2}{6x}$. As all summands are positive (on that interval), we have uniform convergence on any interval $[a,\infty)$ with $a>0$: Given $\epsilon>0$, we certainly have an error $<\epsilon$ for $x>\frac{\pi^2}{6\epsilon}$, hence need only concentrate on the interval $[a,\frac{\pi^2}{6\epsilon}]$, which is compact. In fact, we only used "positive continuous summands with a limit tending to $0$ at infinity". The same trick can be used to show that $f$, with the finitely many summands having $1+n^2a\ge 0$ removed, converges uniformly on $(-\infty,a]$ for any $a<0$. Thus after adding bac those finitely many summands, $f$ converges uniformly on any closed subinterval of $\Bbb R\setminus A$ where $A=\{0\}\cup\{\,-\frac1{n^2}\mid n\in\Bbb N\,\}$ is the set of points where $f$ does not converge (or even some summand is undefined) and tends to infinity.