Proving that $\sum\limits^{\infty}_{n=1}\ln\left(1+\frac{x^2}{n^2}\right)$ converges $\forall x \in ℝ$ to a continuous function

Question

I have tried proving that the series is uniformly continuous $\forall x \in ℝ$ by showing:

$$\lim_{n\to \infty}{\sup_{x\in[-R;R]}{\ln\left(1+\frac{x^2}{n^2}\right)}} = 0$$

But after I did that, I realized that what I need to prove is actually:

$$\lim_{n\to \infty}{\sup_{x\in[-R;R]} \left\lvert\sum\limits^{\infty}_{k=n+1} \ln\left(1+\frac{x^2}{k^2}\right)\right\rvert} = 0$$

and I do not know how to do that, nor am I sure that this is true.

I have also noted that the derivative of $\ln\left(1+\frac{x^2}{n^2}\right)$ is $\frac{2x}{x^2 + n^2}$, but I don't know how to show the uniform convergence of that series, either.

Edit: It was turned to my attention that $0 \le ln(1+y) \le y$, which means that $$\sum\limits^{\infty}_{n=1}\ln\left(1+\frac{x^2}{n^2}\right) \le \sum\limits^{\infty}_{n=1}\frac{x^2}{n^2} = \frac{\pi x^2}{6}$$

But, while certainly showing that the series converges, how does it help me show that it converges uniformly?

Answer 1

Your approach can be completed by noting that $$ \begin{align} \sum_{k=n+1}^\infty\log\left(1+\frac{x^2}{k^2}\right) &=\sum_{k=n+1}^\infty\frac1{k^2}\log\left(\left(1+\frac{x^2}{k^2}\right)^{k^2}\right)\\ &\le x^2\sum_{k=n+1}^\infty\frac1{k^2}\\\\ &\le x^2\sum_{k=n+1}^\infty\frac1{k^2-k}\\[9pt] &=\frac{x^2}n \end{align} $$

Answer 2

This is an overkill, but the theory of Weierstrass products (starting from the Mittag-Leffler theorem) grants that $$\forall x\neq 0,\qquad \frac{\sinh(\pi x)}{\pi x} = \prod_{n\geq 1}\left(1+\frac{x^2}{n^2}\right) $$ holds pointwise and uniformly over any compact subset of $\mathbb{C}$, hence $$\forall x\neq 0,\qquad \sum_{n\geq 1}\log\left(1+\frac{x^2}{n^2}\right)=\log\left(\frac{\sinh (\pi x)}{\pi x}\right)$$ does the same over $\mathbb{R}$, once we re-define $\frac{\sinh(\pi x)}{\pi x}$ at $x=0$ as $1$ (i.e. we remove the removable discontinuity). Then the RHS is clearly continuous.