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?