Uniform convergence of $\sum_{k=1}^{\infty} \frac{1}{k}\arctan(\frac{x}{k})$ on $\mathbb{R}$

Question

$f_n(x) = \displaystyle{\sum_{k=1}^{n} \frac{1}{k}\arctan(\frac{x}{k})}$

Does $f_n(x)$ converge uniformly on $\mathbb{R}$? This sequence of functions clearly converges uniformly on any bounded interval $[-M, M]$, but I wonder whether it converges uniformly on $\mathbb{R}$.

Answer 1

Let's denote $$f(x)=\displaystyle{\sum_{k=1}^{\infty} \frac{1}{k}\arctan(\frac{x}{k})}$$ which is well defined for all $x \in \mathbb R$.

You have $$f(2n)-f_n(2n) \ge \sum_{k=n+1}^{2n} \frac{1}{k}\arctan(\frac{2n}{k}) \ge \frac{\pi}{4} \sum_{k=n+1}^{2n} \frac{1}{k} \ge \frac{\pi}{8}$$

hence the sequence of functions $\{f_n\}$ doesn't converge uniformly towards $f$ which is its pointwise limit.