Prove $$f(x)=\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{n+x^2}$$is differentiable on $(-\infty,+\infty)$
My try: $\Big(\frac{(-1)^{n-1}}{n+x^2}\Big)'=\frac{(-1)^{n}2x}{(n+x^2)^2} $,$\forall a>0 $,
For $x\in[-a,a]$,consider $$g(x)=\sum_{n=1}^{+\infty}\frac{(-1)^{n}2x}{(n+x^2)^2}$$Now $\Big|\frac{(-1)^{n}2x}{(n+x^2)^2}\Big|\le\frac{2a}{n^2}$, hence $\sum_{n=1}^{+\infty}\frac{(-1)^{n}2x}{(n+x^2)^2}$ uniformly converges on $[-a,a]$. Then I conclude $f(x)$ is differentiable on $[-a,a]$. By the arbitrary of $a$, $f(x)$ is differentiable on $[-\infty,+\infty]$. Have I make a rigorous proof?
It is almost correct. However, note that, by the same argument, the series $\sum_{n=1}^\infty1$ defines a differentiable function. You forgot that you should alse check wither your series converges pointwise (it does, by the Weierstrass $M$-test, for instance).