# If $f_n$ is uniformly convergent on each $\{C_n\}_{n\in\mathbb{N}}$, then $f_n$ is uniformly convergent on $\bigcup C_n$?

This is a problem in PMA Rudin p.165

Let $C_m=(-\frac{1}{m^2},-\frac{1}{(m+1)^2}), \forall m\in\mathbb{N}$

Define $f_n=\frac{1}{1+n^2x}$ on its natural domain,$\forall n\in \mathbb{N}$.

Define $f(x)=\sum_{n=1}^{\infty} f_n(x)$ where $x\in \mathbb{R}\setminus (\{0\} \cup \{-\frac{1}{n^2}\in \mathbb{R}|n\in\mathbb{N}\})$.

Then $f$ is well-defined.

I have shown that $\sum f_n\rightarrow f$ uniformly on $C_m$ for any $m\in\mathbb{N}$.

Now, i don't know how to prove that $\sum f_n\rightarrow f$ uniformly on $f$'s natural domain. Is there a theorem for this?

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Looking at my copy of Rudin, I think you want $f_n(x)={1\over 1+n^2x}$. In this case: consider the values of $f_n(1/n^2)$. This should convince you that $(f_n)$ does not converge uniformly to $0$ on $(0,\infty)$. Then the series can't either. (The series does converge uniformly on intervals of the form $[a,\infty)$, $a>0$ by the $M$-test.) – David Mitra Dec 19 '12 at 23:03
I meant the series can't be uniformly convergent on $(0,\infty)$ in my previous comment. – David Mitra Dec 19 '12 at 23:12
@David Oh that's right.. It was a typo – Katlus Dec 20 '12 at 0:02
@copper.hat it's edited now Please check it again. Thank you – Katlus Dec 20 '12 at 0:09
What you are trying to do won't work. Consider the values of $f_n({-1\over2n^2})$ (then argue as in my previous comment). – David Mitra Dec 20 '12 at 0:33