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Show that $$f(x) = \sum_{k=1}^\infty \frac{\arctan(kx)}{k^2}$$ is not differentiable at $x=0$.

By the Weierstrass' test, I can show that $f(x)$ is uniformly convergent and thus continuous. By checking that $f'(x)$ is uniformly convergent if $x \neq 0$, I also know that it's differentiable at those points.

So, $f(x)$ is continuous at $x=0$, but not differentiable. How do I prove that it's not differentiable?


$f'(x) = \sum_{k=1}^\infty \frac{1}{k^3x^2+k}$ and by setting $x=0$, I get $$f'(0) = \sum_{k=1}^\infty \frac{1}{k},$$ which I know diverges. But this is not enough to show that it's not differentiable at $x=0$?

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  • $\begingroup$ Showing that $\lim_{x \to 0} f'(x)$ blows up is not sufficient, you need to show that $\lim_{h \to 0} \frac{1}{h} \sum_k \frac{\arctan(kh)}{k^2}$ doesn't converge. $\endgroup$ – Ian Mar 6 '19 at 23:41
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Notice that $$\begin{align*} \frac{f(h)-f(0)}{h}&=\sum_{k=1}^\infty \frac{\arctan kh}{k^2h}\\&=\sum_{k=1}^\infty \frac{1}{k^2h}\int^{kh}_0 \frac{du}{1+u^2}\\&\ge \sum_{k=1}^\infty \frac{1}{k}\frac{1}{1+k^2h^2}\\&\ge\sum_{k\le \frac 1 h} \frac{1}{2k}=\frac 12 H_{\lfloor \frac 1 h\rfloor}\to \infty \end{align*}$$ as $h\to 0^+$ where $H_n = 1+\frac 1 2+\cdots +\frac 1 n$.

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  • $\begingroup$ Two of your lines are identical. $\endgroup$ – Ian Mar 6 '19 at 23:43
  • $\begingroup$ @Ian Fixed now, thank you! $\endgroup$ – Simon Mar 6 '19 at 23:44
  • $\begingroup$ @Simon How did you go from second line to third line? $\endgroup$ – wednesdaymiko Mar 7 '19 at 1:36
  • $\begingroup$ Noting that $u^2\le k^2h^2$, we get $\int_0^{kh} \frac1{1+u^2}du\ge \int_0^{kh} \frac1{1+k^2h^2}du=\frac{kh}{1+k^2h^2}.$ Hopefully, is it clear now? $\endgroup$ – Simon Mar 7 '19 at 1:39
  • $\begingroup$ @Simon Still a bit confused. Understand the rest, but I'm getting stuck at why and how you arrived at $u^2 \leq k^2h^2$? $\endgroup$ – wednesdaymiko Mar 7 '19 at 2:42

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