Mathematics Asked by Applesauce44 on January 10, 2021
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?
Ishraaq Parvez gave you the answer but I'm going to detail. First the theorem about Alternating Series :
Hypothetis :
Result :
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 = sumlimits^{+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.
Correct answer by math on January 10, 2021
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