every absolutely convergence in interval of convergence is uniformly

Question

I want to prove that if $$\sum_{n=1}^\infty f(x)$$ is absolutely convergence in interval of convergence then its uniformly convergence

I know that if the series of function absolutely convergence then its convergent easy to prove by Cauchy. but what is the effect of interval of convergent to make the series uniformly convergence not only point wise??

Answer 1

There is no such thing... Try $$\sum_{n\geq 0} g_n $$ on $[0,1]$ with $$ g_n:= x^{n}-x^{n+1} \ (\geq 0, \forall x\in [0,1])$$

Indeed, $$\sum_{n= 1}^N g_n(x)= 1-x^{N+1} \underset{N\rightarrow \infty}\longrightarrow g(x) $$ with $g(x)=1$ for $x\in [0,1)$ and $g(1)=0$.

Since $g_n$ is positive, the convergence is absolute but cannot be uniform since if it were, the limit would be continuous.

Answer 2

$$\sum_{n\ge 0}x^n $$ is absolutely convergent at $(-1,1) $ but not uniformly.

$$S (x)-S_n (x)=\frac{x^n}{1-x} $$

is not bounded.