Uniform convergence of series of functions: $\sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{x^2+\sqrt{k}}$

Question

I have a question about the uniform convergence of the following series :

$$\sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{x^2+\sqrt{k}}$$

So far I know that the series is convergent but not absolutely convergent.

I am missing the last step(s) of the following tasks: $$\forall n\in \mathbb{N}~\setminus \{0\}, ~f_n(x):= \sum^{n}_{k=1}\frac{(-1)^{k+1}}{x^2+\sqrt{k}} \forall x\in \mathbb{R}$$ Show that:

(i) $$\Vert f_{n+2p}-f_n \Vert_{\infty,\mathbb{R}} \leq \sum^{p}_{k=1}\frac{1}{(n+2k-1)^{3/2}} ~\forall p \in \mathbb{N}\setminus \{0\}$$

(ii)$$\Vert f_{n+2p+1}-f_n \Vert_{\infty,\mathbb{R}} \leq \Vert f_{n+2p}-f_n \Vert_{\infty,\mathbb{R}}+(n+1)^{-1/2}~ \forall p \in \mathbb{N}\setminus \{0\}$$

(iii) Hence, conclude that the series $$\sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{x^2+\sqrt{k}}$$ is uniformly convergent on $\mathbb{R}$.

I came as far as this:

(i) $$\Vert f_{n+2p}-f_n \Vert_{\infty,\mathbb{R}}=\sup_{x\in\mathbb{R}}\Vert\sum^{n+2p}_{k=n+1}\frac{(-1)^{k+1}}{x^2+\sqrt{k}}\Vert=\sup_{x\in\mathbb{R}}\sum^{n+2p}_{k=n+1}\frac{1}{x^2+\sqrt{k}}=\sum^{n+2p}_{k=n+1}\frac{1}{\sqrt{k}} $$ ...

My idea is to somehow change indices and I really dont know how to get the $n$ and the $(3/2)$ power in the series. I am thankful for any help.

Cheers.

Answer 1

Hint: an alternating series $\sum (-1)^k a_k$, with $(a_k)$ sequence of non-negative numbers monotonically decreasing to $0$, is convergent (Leibniz criterion). Calling $s$ the sum of the series and $s_n := \sum_{k=1}^n (-1)^k a_k$, it holds $$ |s - s_n| \leq a_{n+1}, \qquad \forall n\in\mathbb{N}. $$

Answer 2

Write $$f_{n+2p}-f_n=\sum_{k=0}^{p-1} (g_{n+2k+1}+g_{n+2k+2}).$$ Then \begin{align}g_{n+2k+1}(x)+g_{n+2k+2}(x)=\frac{1}{x^2+\sqrt{n+2k+1}}-\frac{1}{x^2+\sqrt{n+2k+2}}\\ =\frac{\sqrt{n+2k+2}-\sqrt{n+2k+1}}{(x^2+\sqrt{n+2k+1})(x^2+\sqrt{n+2k+2})} \\=\frac{2}{(\sqrt{n+2k+2}+\sqrt{n+2k+1})(x^2+\sqrt{n+2k+1})(x^2+\sqrt{n+2k+2})}\end{align} which should give you the power $\frac32$.