Uniform convergence of $1/(1+n^2x)$ of course, we know that $\frac{1}{1+n^2x}$ converges uniformly on $[a,\infty] $ for a>0; for x>0.

I am having difficulty in proving for $x&lt;0$. We have discontinuities at $x=1/k^2, \forall k\in Z^-.$ So for now I just wanted to prove it for $x&lt;-1$

This is my attempt, if you could check it.

$$|1+n^2x|=|1-n^2|x||&gt;n|1-n|x||$$so

$$\bigg|\frac{1}{1+n^2x}\bigg|&lt;\bigg|\frac{1}{n(n|x|-1)}\bigg|&lt;\bigg|\frac{1}{n(n-1)}\bigg| $$

and so the series converges uniformly for $x&lt;-1$

Further, what do I do for $-1&lt;x&lt;0$ . Thank you
If a&lt;0 ,then there exist a K such that for all n>K ,a>-1/n^2.At -1/n^2 ,function tends to infinity.This is true for all n>K.So sequence is not convergent in sup norm .So it isn't uniformily convergent
If a&lt;0 ,then there exist a K such that for all n>K ,a>-1/n^2.At -1/n^2 ,function tends to infinity.This is true for all n>K.So sequence is not convergent in sup norm .So it isn't uniformily convergent