Uniform convergence of $\sum \frac{nx}{1+n^2x^2}$ I know that $\frac{nx}{1+n^2x^2}$ isn't uniformly convergent.
However I am not sure about the series. None of the tests for series convergence, namely wierstrass m test, Abel test or Dirichlet test give anything about proving non convergence.
Does non-uniform convergence of individual terms imply non-uniform convergence of series?
It's overkill in this case since the series
$$
\sum_{n=1}^\infty \frac{nx}{1+n^2x^2}
$$
is not even pointwise convergent for $x\neq 0$, but in general:

If you have uniform convergence over $E$ of a series with partial sum $S_n=\sum_{k=1}^n f_n$ to a limit function $S$, then
$$\begin{align*}
\sup_{x\in E}|f_n(x)| &amp;= \sup_{x\in E} |S_{n}(x)-S_{n-1}(x)| \leq \sup_{x\in E} |S_{n}(x)-S(x)|+ \sup_{x\in E} |S_{n-1}(x)-S(x)| \\&amp;\xrightarrow[n\to\infty]{} 0+0=0
\end{align*}$$
so $(f_n)_n$ must converge uniformly (on $E$) to the zero function.
Let $f_n(x)=\frac{n x}{1+n^2 x}$, the series $\sum_{n=1}^{\infty} \frac{n x}{1+n^2 x}$ is not defined on $x=\frac{-1}{n^2},\,\, n=1,2,3,...$. and for all other values of $x$, we have $\frac{n x}{1+n^2 x} \to 0,\,\, as\,\, n \to 0$.
According to M-test theorem, the series convergent uniformly on any bounded interval $(a,b],\,\, b &gt;a&gt;0$ for
$$\frac{n x}{1+n^2 x} \le \frac{b} {a n^2} $$
and it is clear that
$$\frac{b}{a}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{b \pi^2}{6 a}.$$
The series diverges at $x= \frac{1}{1-n^2},\,\,\, n=1,2,3,...$ because the expression $\frac{n x}{1+n^2 x} =1$ at $x=\frac{1}{1-n^2}$ and the infinite summation for the last diverges.