Let $f_n: [0, \infty) \rightarrow \mathbb{R}$ be a sequence of real functions.
Suppose for each $n$ and $x \in [0,\infty)$ we have that $0 \leq f_{n+1}(x) \leq f_n(x)$.
Show that if $(M_n)_{n=1}^{\infty}$ is a sequence of reals with $\lim_{n\to\infty} M_n = 0$ and $\forall x \in [0, \infty) \space \space \forall n \in \mathbb{N}:\space 0 \leq f_{n}(x) \leq M_n$
then $\sum(-1)^{k}f_{k}(x)$ converges uniformly over $[0, \infty)$.
I know, by Leibniz's test, that the sequence convergence pointwisely.
However it's left unclear to me why the existence of $(M_n)$ would ensure that the convergence is uniformly. I will be glad to have a hint\an observation which I possibly haven't considered yet.
Thanks
Let $S_{n}(x)=\sum_{k=1}^{n}(-1)^{k}f_{k}(x)$ be the partial sum. We go to show that $(S_{n}(x))_{n}$ is uniformly Cauchy and hence it converges uniformly. Let $\varepsilon>0$ be given. Choose $N$ such that $M_{n}<\varepsilon$ whenever $n\geq N$. Let $p\in\mathbb{N}$, $n\geq N$, $x\in[0,\infty)$ be arbitrary, then \begin{eqnarray*} & & \left|S_{n+p}(x)-S_{n}(x)\right|\\ & \leq & \left|(-1)^{n+1}f_{n+1}(x)\right|\\ & \leq & M_{n+1}\\ & < & \varepsilon \end{eqnarray*} This shows that $(S_{n}(x))_{n}$ is uniformly Cauchy on $[0,\infty)$. In the above, we have used the fact that $f_{1}(x)\geq f_{2}(x)\geq\ldots\geq0$ to infer that $\left|S_{n+p}(x)-S_{n}(x)\right|\leq\left|(-1)^{n+1}f_{n+1}(x)\right|$.