## 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 $$x=0$$?

## If $\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} f_{n}(k) = \infty$ then $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} f_{n}(k) = \infty$.

Suppose $$\{f_{n}\}_{n=1}^{\infty}$$ be functions such that $$f_{n} : \Bbb{N} \rightarrow \Bbb{R}^{+}$$ for each $$n$$.

I was trying to prove –

If $$\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} f_{n}(k) = \infty$$ then $$\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} f_{n}(k) = \infty$$.

I can see that both the double summation series are equal by expanding the double summation.But I am trying to prove it ? any other thoughts?

Hm, as the order of summation are changed, is Uniform convergence likely to play any role here?