How to prove that this series of functions is uniformly convergent?

Question

Let $\{f_n\}$ be a sequence of functions on $[a, b]$ such that: $f_n(x) \le 0$ if $n$ is even, $f_n(x) \ge 0$ if $n$ is odd; $|f_n(x)| \geq |f_{n+1}(x)|$ for all $x$; $f_n$ converges to $0$ uniformly.

I have to prove that the summation of $f_n$ is uniformly convergent.

How do I approach this problem? Where should I start?

Answer 1

Ishraaq Parvez gave you the answer but I'm going to detail. First the theorem about Alternating Series :

Hypothetis :

• $(u_n)$ is alternate, which means either $((-1)^n u_n)_{n\in \mathbb{N}}$ or $((-1)^{n+1} u_n)_{n\in \mathbb{N}}$ is positive
• $(|u_n|)$ is decreasing
• $|u_n|\rightarrow 0$ when $n\rightarrow +\infty$

Result :

• $\sum u_n$ converges
• if we note $R_n = \sum\limits_{i=n+1}{+\infty} u_i$, then for all $n\in \mathbb{N}$, $|R_n|\leqslant |u_n|$

Then for your problem : it is clear that for all $x$, $(f_n(x))$ is alternate. We also know that $\sum f_n$ uniformly convergent is equivalent to $R_n = \sum\limits^{+\infty}_{i=n+1} f_i $ uniformly converges to $0$. By the theorem on alternate series, we have first that $\sum f_n$ exists, and also that for all $x$, $|R_n|\leqslant |f_n(x)|$. So we have that $\sup |R_n| \leqslant \sup |f_n|$. (you will have remarked that the sup are defined since we have convergence)

Since $(f_n)$ uniformly converges to $0$, $\sup |f_n| \rightarrow 0$. So $\sup |R_n| \rightarrow 0$. Thus $\sum f_n$ uniformly converges.