Showing Uniform Continuity of An Infinite Series and Termwise Differentiation This is a problem from Kenneth A Ross 2nd Edition Elementary Analysis:
Show that the infinite series,$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n+x^2}$$ converges uniformly for all $x$, and by termwise differentiation, compute $f '(x)$.
My work so far involves the Weierstrass M Test - essentially $\lvert\frac{(-1)^n}{n+x^2}\rvert \leq \lvert\frac{1}{k^2}\rvert \forall x \in \mathbb{R}$ and since $\sum_{n=1}^{\infty} \frac{1}{k^2} \lt \infty$, the original series must be uniformly convergent for all x.
I'm not exactly sure if I did that right, or if the alternating negative one changes the requirements for the M-test.
Edit - based on comments, the alternating series test seems useful here. Because of the $(-1)^n$ component, and for each term $a_n$ in the series $a_{n+1} \leq a_n$, that means the series converges by the alternating series test. But does that show absolute convergence, and is that enough, or does it need more?
 A: HINT: Let $f_k(x)=\displaystyle{\sum_{n=1}^k \frac{(-1)^n}{x^2+n}}$. You know that there is $f(x)$ so that $f_k(x)\to f(x)$ for each $x$. What is an estimate on $|f(x)-f_k(x)|$? Can you get that estimate independent of $x$?
A: Uniform convergence of $f(x) = \sum_{n=1}^{\infty} \frac{(-1)^n}{n+x^2}$ on $\mathbb{R}$ follows from the Dirichlet test. Another way to prove this is to observe that for all $x \in \mathbb{R}$,
$$\left|\sum_{k=n+1}^{m} \frac{(-1)^n}{k+x^2}\right| = \frac{1}{n+1+x^2}- \left(\frac{1}{n+2+x^2}- \frac{1}{n+3+x^2} \right) - \ldots \\\leqslant \frac{1}{n+1+x^2} \leqslant \frac{1}{n+1}$$
and uniform convergence follows by the uniform Cauchy criterion.
The termwise differentiated series is
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}2x}{(n+x^2)^2}$$
We have
$$\left|\frac{(-1)^{n+1}2x}{(n+x^2)^2}\right| = \frac{1}{\sqrt{n}}\frac{1}{n+|x|^2}\frac{2\sqrt{n}|x|}{n+|x|^2} \leqslant \frac{1}{n^{3/2}},$$
and this series converges uniformly by the Weierstrass M-test.
By a well-known theorem it follows that $f$ is differentiable and $f'(x)$ equals the sum of the termwise differentiated series.
