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Well , this is the function sequence:$$f_k\left(x\right)\:=\:\frac{1}{k+k^2x}$$
I want to prove that there is no uniform convergence for $\sum _{k=1}^{\infty }\:\frac{1}{k+k^2x}$ , in ($0$,$\infty $).

I thought about show the sequence is not uniformly converge in ($0$,$\infty $) but i don't know how can i show it. How can i do it?

Edit: the functions are all from ($0$,$\infty $) to R

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    $\begingroup$ you need to start the sum at k=1 so you aren't dividing by 0. $\endgroup$ – rVitale Jun 24 '14 at 14:43
  • $\begingroup$ ok i edited this $\endgroup$ – user2637293 Jun 24 '14 at 14:46
  • $\begingroup$ Show that for any $N$, there is an $L$ so that $\lim_{x\rightarrow0+}\sum_{k=N}^{N+L} f_k(x)$ is greater than $1$. $\endgroup$ – David Mitra Jun 24 '14 at 14:48
  • $\begingroup$ As a source of many examples, take just about any power series with infinite radius of convergence. $\endgroup$ – Andrés E. Caicedo Jun 24 '14 at 14:56
  • $\begingroup$ @DavidMitra I see u using Cauchy. so i started and i get $\left|\sum \:_{k=m+1}^{n\:}\:\frac{1}{k+k^2x}\right|\:<\left|\:\sum \:_{k=m+1}^n\:\frac{1}{k^2x}\right|$. what i need to do from here? $\endgroup$ – user2637293 Jun 24 '14 at 14:57
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Note that for any $n$

$$\sup_{x \in (0,\infty)}\sum_{k=n+1}^{\infty}\frac{1}{k+k^2x}>\sup_{x \in (0,\infty)}\sum_{k=n+1}^{2n}\frac{1}{k+k^2x}>\sup_{x \in (0,\infty)}\frac{n}{2n+4n^2x}> \frac1{2},$$

so the convergence is not uniform on $(0,\infty)$

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