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I have series $$\sum_{n=0}^{\infty} 1/(1 + n^2x^2) $$ for $x \in (0,1]$. I think this will not converge uniformly if we pick $x = 1/n$ since $x \in (0,1]$ and we can see that this will contradict the original definition for uniform convergence because we can see for interval $(0,1]$ the point wise limit is $0$ so picking $x = 1/n$ we and choosing $\varepsilon = 1/2$

$|\sum 1/(1 + 1)| $ = $\sum 1/2$ > 1/2 so it will not uniformly converge. What is wrong with this argument?

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You cannot choose $x$ dependigng on $n$. Taking $x=1/n$ only the $n$-th term is equal to $1/2$.

The series converges uniformly on any interval $[a,1]$ with $0<a<1$ by the Weierstrass $M$-test: $$ \frac{1}{1+n^2\,x^2}\le\frac{1}{n^2\,x^2}\le\frac{1}{a^2\,n^2}. $$ But it does not converge uniformly on $(0,1]$, since it does not satisfy the uniform Cauchy criterion. If $m<n$ $$ \sup_{0<x\le1}\sum_{k=m}^{n}\frac{1}{1+n^2\,x^2}\ge\sup_{0<x\le1}\frac{1}{1+n^2\,x^2}\ge\frac12. $$

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