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

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 $$x=0$$?

• 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. – Ian Mar 6 '19 at 23:41

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 $$H_n = 1+\frac 1 2+\cdots +\frac 1 n$$.
• 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? – Simon Mar 7 '19 at 1:39
• @Simon Still a bit confused. Understand the rest, but I'm getting stuck at why and how you arrived at $u^2 \leq k^2h^2$? – wednesdaymiko Mar 7 '19 at 2:42