Show that $\sum a_nx^n$ converges. The problem is: Suppose the convergence radius of the power series $\sum a_n x^n$ is equal to $1$, also the coefficients of the series satisfy $a_1\geq a_2\geq a_3\geq ...$ and $a_n\to 0$ when $n\to\infty$. Show that $\sum a_n x^n$ converges for all $x$ such that $|x|=1$, except maybe at $x=1$.
My attempt: Let $b_n=a_nx^n$, if we take $$\displaystyle\lim_{n\to\infty}\bigg\vert\frac{b_{n+1}}{b_n}\bigg\vert=\lim_{n\to\infty}\bigg\vert\frac{a_{n+1}}{a_n}\bigg\vert\vert x\vert=0$$ if $|x|\leq 1$, then, by the ratio test, the serie $\sum a_nx^n$ converges for all $|x|=1$. Observe that, the series $\sum\frac{x^n}{n}$ when $x=1$ is the harmonic series, and does not converge, but when $x=-1$ it does, then the series does not necessarily converge when $x=1$.
I know that my attempt is wrong since I don't use the fact that the convergence radius is 1 or that $a_n\to 0$, but I don't have any ideas on how to fix it or do it.
Thanks!!!
 A: This question is tagged real-analysis, so I assume by $|x|=1$ you mean $x\in\{-1,1\}$. The convergence for $x=-1$ is given by Dirichlet's Test which says
$\require{cancel}$

If $\{a_n\}$ is a sequence of real numbers and $\{b_n\}$ is a sequence of $\overset{\large\text{real}}{\cancel{\text{complex}}}$ numbers satisfying

*

*$a_n\ge a_{n+1}$

*$\lim\limits_{n\to\infty}a_n=0$

*$\displaystyle\left|\ \sum_{n=1}^Nb_n\ \right|\le M$ for every positive integer $N$
where $M$ is some constant, then the series
$\displaystyle\sum_{n=1}^\infty a_nb_n$
converges.

Since $\displaystyle\sum_{n=1}^N(-1)^{n-1}\in\{0,1\}$, Dirichlet's Test applies.
A: Since you tagged your question real analysis, I'll assume you are concerned with $x \in [-1,1]$.
As you mentioned, the harmonic series is the counterexample for $x=1$.
For $|x|<1$ note that:
$$
|a_n x^n| \leq |x|^n \text{ and that } \sum_{n=1}^\infty |x|^n < \infty
$$ since the latter is a geometric sum with rate $|x|$.
For the case $x=-1$, this is the alternating harmonic series and is often an example of a series which is convergent but not absolutely convergent.
