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
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 $(0,\infty)$