-1
$\begingroup$

How do I prove that $\sum_ {k=0}^ {\infty} \frac{\sin (kx)^2}{ (1+k^ 2x^2)}$ is not continuous at zero?

This is a question on uniform convergence I tried to solve it I just don't have any clue how I can approach this problem. Can you give me any hints?

$\endgroup$
2
  • 1
    $\begingroup$ Have you tried evaluating this at $0$? $\endgroup$
    – AnCar
    Commented Apr 15, 2022 at 12:17
  • $\begingroup$ @AnCar yeah it's zero i just don't know how to take the limit $\endgroup$ Commented Apr 15, 2022 at 12:21

1 Answer 1

2
$\begingroup$

Let us call $f(x):=\sum_{k=0}^{+\infty}\frac{\sin^2(kx)}{1+k^2x^2}$. Note that $f(0)=0$ and that continuity would imply $f(x_n)\to 0$ as $n\to +\infty$ for every sequence $(x_n)_n$ converging to $0$. However you can see that the choice $x_n=\frac{\pi}{2n}$ gives $f(x_n)\geq\frac{1}{1+\frac{\pi^2}{4}}$ for every $n$, thus $f(x_n)$ cannot converge to $0$.

$\endgroup$
3
  • $\begingroup$ how did you get that lower bound for f(x_n)? $\endgroup$ Commented Apr 15, 2022 at 12:42
  • $\begingroup$ The terms of the sum are all non-negative. Think what happens when you look at the term with $k=n$. $\endgroup$
    – AnCar
    Commented Apr 15, 2022 at 12:49
  • $\begingroup$ @AnCar Got it, Thanks $\endgroup$ Commented Apr 15, 2022 at 17:34

Not the answer you're looking for? Browse other questions tagged .