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

Question

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

1. $\mathbb{R}$
2. $[a,b]$ (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?

Answer 1

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 $x=n$.

Answer 2

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 $[a,b]$