$f(x)$ is a continuos fuction in $[0,1]$ and differentiable twice on $0$.
$U_n=(-1)^{n}f(\frac{1}{n})$ I need to prove that:
$1.$ if $f(0)=0$ then $\sum_{n=1}^\infty U_n$ converge.
$2.$ if $\sum_{n=1}^\infty U_n$ converge absolutely then $f(0)=f'(0)=0.$
I used Maclaurin series with 2 degree: $f\left(\frac{1}{n}\right) = \frac{f'(0)}{n}+\frac{f''(0)}{2n^{2}} + o\left(\frac{1}{n^2}\right)$.
I can't proceed from here can you help me? Thanks.