Uniform convergence of $\sum_{n=1}^{\infty}\ln\left(1+\frac{x^2}{n^2}\right)$ Find the convergence domain and determine if $\sum_{n=1}^{\infty}\ln(1+\frac{x^2}{n^2})$ converges uniformly on


*

*$\mathbb{R}$

*$[a,b]$ (some closed interval)


An Attempt:
Using $\ln(1+t)\leq t$, we will have:
$$\sum_{n=1}^{\infty}\ln\left(1+\frac{x^2}{n^2}\right)<\sum_{n=1}^{\infty}\frac{x^2}{n^2}=x^2\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$$
Thus, the series converges for every $x\in\mathbb{R}$.
For 2: We can use M-test and get  $\sum_{n=1}^{\infty}\ln\left(1+\frac{x^2}{n^2}\right)<b^2\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$  
What I need to do with 1? 
 A: If $\sum\limits_{n=1}^\infty f_n(x)$ is uniformly convergent on $I$, then it is uniformly Cauchy on $I$.  That is, for each $\epsilon>0$, there is an $N$ so that for all $m\ge n\ge N$, we have 
$$
\Bigr| \sum_{i=n}^m f_i(x) \Bigl|<\epsilon, \ \text{for all}\ x\in I.
$$
See this post for a proof of this.
From the above, it follows that if $\sum\limits_{n=1}^\infty f_n(x)$ is uniformly convergent on $I$, then the sequence $(f_n)$ converges to $0$ uniformly on $I$.
The terms of your series do not converge to $0$ uniformly on $\Bbb R$. To see this, evaluate the $n$'th term of the series at $x=n$.
A: Let 
$$R_n(x)=\sum_{k=n+1}^\infty \ln\left(1+\frac{x^2}{k^2}\right)$$
the remainder of the series then we have 
$$R_n(x)\sim_\infty x^2 \sum_{k=n+1}^\infty \frac{1}{k^2}\sim_\infty x^2\int_{n+1}^\infty\frac{dt}{t^2}\sim_\infty \frac{x^2}{n}$$
hence for $x_n=\sqrt{n}$ we have 
$$\lim_{n\to\infty}R_n(x_n)=1\neq 0$$
and the series doesn't converge unifomly on $\mathbb{R}$ and 
$$\sup_{x\in[a,b]} |R_n(x)|\sim_\infty\frac{M}{n^2}\to0$$
so it converges uniformy in every closed intervel $[a,b]$
