1
$\begingroup$

I have to prove that if $\sum\limits_{n=1}^\infty \frac{1}{|a_n|}$ converges, then $\sum\limits_{n=1}^\infty \frac{1}{x-a_n}$ converges absolutely and uniformly in any compact space that $\forall n \in \mathbb N$ doesn't contain point $x=a_n$.

Since there is a compact space and all functions $\frac{1}{x-a_n}$ are continuous on it, maybe it's worth to use Dini's theorem, but I can't figure out how to apply it, because I need to show that this series converges pointwise in this compact space.

$\endgroup$
  • 2
    $\begingroup$ Not Dini, you don't know whether the convergence is monotonic (and if the $a_n$ are complex numbers, monotonic generally doesn't even make sense). Use the Weierstraß test. Can you find a lower bound for $\lvert x-a_n\rvert$ in terms of $\lvert a_n\rvert$ for large $n$? $\endgroup$ – Daniel Fischer Oct 20 '14 at 20:02
  • $\begingroup$ I just see that $|x - a_n| \ge ||x| - |a_n||$ $\endgroup$ – perlik Oct 20 '14 at 20:20
  • 1
    $\begingroup$ Good start. Now, for large $n$, you know that $\lvert a_n\rvert$ also must be large, otherwise $\sum \frac{1}{\lvert a_n\rvert}$ wouldn't be finite. So, since $x$ lies in some compact set $K$, what can you say about $\lvert x\rvert$? $\endgroup$ – Daniel Fischer Oct 20 '14 at 20:23
  • $\begingroup$ We can say that $A \le |x| \le B$ for some $A,B$. $\endgroup$ – perlik Oct 20 '14 at 20:44
  • 1
    $\begingroup$ We only need $\lvert x\rvert \leqslant B$. Now, is there an $n_0$ such that for $n\geqslant n_0$ you have $\lvert a_n\rvert \geqslant 2B$, say? $\endgroup$ – Daniel Fischer Oct 20 '14 at 20:52

Your Answer

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

Browse other questions tagged or ask your own question.