Show that $\displaystyle\sum_{n=1}^\infty \frac{x}{1+n^2x}$ is uniformly convergent in [ $\delta,1]$ for any $\delta>0$ but is not uniformly convergent in $[0, 1]$.
I tried like this: Let $S_n=\sum_{n=1}^n f_n=\displaystyle\sum_{n=1}^n\frac{x}{1+n^2x}$ be the $n$th partial sum. Then for any interval in [0,1], we have: $\displaystyle\frac{d}{dx}f_n(x)=\frac{d}{dx}\frac{x}{1+n^2x}=\frac{1}{(1+n^2x)^2}>0, \forall x\in[0,1]$.
Thus $f_n(x)$ being an increasing function, the maximum occurs at x=1. So, $|f_n(x)|\leq|\frac{x}{1+n^2x}|_{x=1}=\frac{1}{1+n^2}<\frac{1}{n^2}$, which is convergent by $p-$test. Hence, by Weierstrass’s M-test, the series is uniformly convergent in any interval with left end point$\geq$0 and right end point as 1.
But this would mean that it is uniformly convergent in [0,1] also, which contradicts the question. Pls help. Thank you.