Uniform convergence of $\sum_{n\geq1} \ln \left(1+\frac{|x|}{n^{2}}\right)$

Question

Does the series of functions $$ \sum_{n\geq1} \ln \left(1+\frac{|x|}{n^{2}}\right) $$ converge uniformly on $\mathbb{R}$?

What is known:

• The series converges pointwise to $f:\mathbb{R}\mapsto \mathbb{R}$, where $f(x)=\ln \left(\frac{\mathrm{e}^{\pi \sqrt{|\mathrm{x}|}}-\mathrm{e}^{-\pi \sqrt{| \mathrm{x}} |}}{2 \pi \sqrt{|\mathrm{x}|}}\right)$ for $x\neq 0$ and $f(0)=0$. This has been found out using the forumla $\sin (\pi z)=\pi z \prod_{n\geq 1}\left(1-\frac{z^{2}}{n^{2}}\right)$ for $z \in \mathbb{C}$. $f$ is continuous on $\mathbb{R}$.
• The series converges uniformly on any bounded interval of $\mathbb{R}$. This can be shown by seeing that for a bounded interval $I\subseteq (-r,r)$ $\left(r \in \mathbb{R}_{>0}\right)$, one has $\forall x\in I: 0 \leq \ln \left(1+\frac{|x|}{n^{2}}\right)\leq \frac{|x|}{n^{2}}\leq \frac{r}{n^{2}}$, and $\sum_{n\geq1}\frac{r}{n^{2}}$ converges. The Weierstrass criterion implies uniform convergence on $I$ of the series of functions we're investigating.

Answer 1

The series does not converge uniformly on $\mathbb R$ because $ln (1+\frac {|x|} {n^{2}}) $ does not tend to $0$ uniformly. The supremum of thus is at least $\ln 2$ as seen by by putting $x=n^{2}$.