Weierstrass test on $\frac{1}{1+n^2x}$ on $(0,1)$ I have shown that $\sum\limits_{n=1}^{\infty}\frac{1}{1+n^2x}$ converges pointwise on $(0,1)$ by fixing $x$ and showing that each series at every $x$ converges using the comparison test.
Now to prove uniform convergence, I am trying to use the Weierstrass test by I cannot come up with a good $M_n$ that does not depend on $x$. Can someone point me in the right direction?
 A: One can prove uniform convergence on any $[a,+\infty[$ with $0<a <+\infty$ with little knowledge using Cauchy criterium. Ideed, for such interval and $0<N<M$, one has 
$$
\left|\sum\limits_{n=N}^{M}\frac{1}{1+n^2x}\right|\leq \left|\sum\limits_{n=N}^{M}\frac{1}{1+n^2a}\right|\leq \left|\sum\limits_{n=N}^{M}\frac{1}{n^2a}\right|\leq \frac{1}{a}\sum\limits_{n=N}^{M}\frac{1}{n^2}\leq \frac{1}{a}\sum\limits_{n=N}^{\infty}\frac{1}{n^2}  
$$
which tends to zero as $N\rightarrow \infty$ as 
$$
\sum\limits_{n=N}^{\infty}\frac{1}{n^2}\leq \int_{N-1}^\infty \frac{dt}{t^2}=\frac{1}{(N-1)}   
$$
All this can be adapted to Weierstrass $M$-test easily as the functions are positive (as I am french Cauchy comes first to the mind, sorry) as follows 
$$
\sum\limits_{n=1}^{N}\left|\frac{1}{1+n^2x}\right|\leq  \frac{1}{a}\left(\sum\limits_{n=1}^{N}\frac{1}{n^2}\right)\leq \frac{1}{a}\left(1+\sum\limits_{n=2}^{N}\frac{1}{n^2}\right)\leq$$
$$ \frac{1}{a}\left(1+\int_{1}^\infty \frac{dt}{t^2}\right)=\frac{2}{a}  
$$
A: $$ 0<x<1  \rightarrow 1+n^2x>1 \rightarrow 0<\frac{1}{1+n^2x}<1\\\frac{1}{1+n^2x}<\frac{1}{n^2x}$$ $$\sum_{1}^{\infty} \frac{1}{1+n^2x} <\sum_{1}^{\infty} \frac{1}{n^2x}=\frac {1}{x}\sum_{1}^{\infty} \frac{1}{n^2}=\frac {1}{x} \frac{\pi^2}{6} $$
A: Suppose that the sequence of functions converges uniformly on $(0,1)$. Then for any $\epsilon\gt0$, we can find an $N$ so that for $x\in(0,1)$ and $n\ge N$, we have
$$
\left|\frac1{1+n^2x}-0\right|\le\epsilon
$$
This would mean that $1\le\epsilon(1+N^2x)$, which in turn means $x\ge\frac{1-\epsilon}{\epsilon N^2}$. However, this is not true for $x\in\left(0,\frac{1-\epsilon}{\epsilon N^2}\right)$. Contradiction.
Therefore, this sequence of functions does not converge uniformly on $(0,1)$.
