Prove convergence of the power series $\sum_{n=1}^\infty \frac{x^n}{\log(n+2)}$

Question

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!

Answer 1

At $x=-1$ you get an alternating series $\sum_n(-1)^n\frac{1}{\log(n+2)}$. Since $\frac{1}{\log(n+2)}\searrow0$ as $n\rightarrow\infty$, the series converges (conditionally).

At $x=1$, since $0<\log(n+2)\leq n+2$, and $\sum_n\frac{1}{n+2}$ diverges to infinity, we get that $\sum_n\frac{1}{\log(n+2)}$ diverges to infinity too.

Answer 2

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 $[-1,1)$.