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

  • 1
    $\begingroup$ 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

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 $(0,\infty)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.