I need to prove the convergence (pointwise, absolute and uniform) of the power serie $\sum_{n=1}^\infty \frac{x^n}{\log(n+2)}$, what I know for now is using the Radio test with $c_n = \frac{1}{\log(n+2)}$ the result is $R=1$, so the series converge absolute (so pointwise) in $(-1,1)$ and uniformly in any compact $K \subset (-1,1)$, and there in no convergence in any apoint $\mathbb{R}\backslash[-1,1]$.
So what about the $-1$ and $1$ points? Is uniform convergence in $(-1,1)$? In the case, $-1$ and $1$ the series converge because it tends to zero, but I can' figure out another series to prove the uniform convergence with the Weierstrass-M test. Any help is appreciated!
Best Answer
That series diverges when $x=1$ (since $n>1\implies\frac1{\log(n+2)}\geqslant\frac1n$) and converges when $x=-1$ (by the alternating series test).
Since it converges when $x=-1$, it follows from Abel's theorem that it converges uniformly on $[-1,0]$. Actually, it converges uniformly in $[-1,r]$ for any $r\in(0,1)$, but it does not converge uniformly in $[-1,1)$.