Show that the series $\sum\limits_{k=1}^{\infty}\dfrac{(-1)^k}{k+|x|}$ converges pointwise to a Lipschitz function $f(x)$. Is the convergence uniform on $\mathbb{R}$?
I have shown the pointwise convergence part. But I have facing trouble in showing the other two parts, that is, uniform convergence part and Lipschitz part
Since the sequence $(k + |x|)^{-1}$ is monotone decreasing for fixed $x$ and uniformly convergent to $0$ , uniform convergence of the series on $\mathbb{R}$ follows from the Dirichlet test or more directly from the alternating series estimate
$$\left|\sum_{k=n+1}^m \frac{(-1)^k}{k+|x|}\right|\leqslant \frac{1}{n+1+|x|}\leqslant \frac{1}{n+1}$$
$$\left|\sum_{k=n+1}^m(-1)^k a_k \right|= \begin{cases} a_{n+1} - (a_{n+2} - a_{n+3}) - \ldots -(a_{m-1} - a_m),,, m-n \text{ odd} \a_{n+1} - (a_{n+2} - a_{n+3}) - \ldots -(a_{m-2} - a_{m-1})- a_m, ,,m-n \text{ even}\end{cases}< a_{n+1}$$
– RRL Oct 27 '23 at 16:21The Lipschitz property part can be shown in a straight forward fashion: $$|f(x) - f(y)|< |\sum_{k=1}^{\infty}\frac{(-1)^k}{k+|x|} - \sum_{k=1}^{\infty}\frac{(-1)^k}{k+|y|}| = |\sum_{k=1}^{\infty}\frac{(-1)^k(|x| - |y|)}{(k+|x|)(k+|y|)}|= |\sum_{k=1}^{\infty}\frac{(-1)^k}{(k+|x|)(k+|y|)}|\cdot ||x| - |y|| \leq |\sum_{k=1}^{\infty}\frac{(-1)^k}{(k+|x|)(k+|y|)}||\cdot|x-y|$$
Set $K: = \sum_{k=1}^{\infty}\frac{(-1)^k}{(k+|x|)(k+|y|)} < \sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{\pi^2}{6}$. Thus, $f$ is a Lipschitz function.
Note that by this argument $f(x) := \sum_{k=1}^{n}\frac{(-1)^k}{k+|x|}$ is a Lipschitz function, and a sequence of $K$− Lipschitz functions which converges pointwise, converge uniformly (see Given sequence of $L-$Lipschitz functions which converges pointwise, prove uniform convergence).
