Finding all points $x \in \mathbb{R}$ such that $\sum_{n=0}^{\infty} a_nx^n$ converges For the series
$\sum_{n=0}^{\infty} a_nx^n = 1 + 2x + x^2+2x^3 + \dotsc$ where  $a_n =
            \begin{cases}
            1,  & \text{if $n$ is even} \\[2ex]
            2, & \text{if $n$ is odd}
            \end{cases}$
find all points $x \in \mathbb{R}$ such that the sum converges.
I just got this question on a test and was limited on time so I made up some answer that is more than likely wrong. I'm just curious on how wrong.
My thought process was $\sum_{n=0}^{\infty} x^n \leq \sum_{n=0}^{\infty} |a_nx^n| \leq \sum_{n=0}^{\infty} 2x^n$ which both converge for $|x| \lt 1$
so the orginal series has the same radius meaning it converges for all $x \in  \left( -1,1\right)$. So how bad is this or is it somewhat reasonable?
 A: The series is $$(1+x+x^2+ \dots) + (x^2+x^4+ \dots)=\frac{1}{1-x} + \frac{x^2}{1-x^2},$$ by the standard geometric series.  This converges for $|x| < 1$.  When $|x|\geq 1$, the series obviously doesn't converge because the terms don't tend to 0.
A: Instead of $\sum_{n=0}^{\infty} x^n \leq \sum_{n=0}^{\infty} |a_nx^n| \leq \sum_{n=0}^{\infty} 2x^n$ for $|x|<1$, you should write
$\sum_{n=0}^{\infty} |x^n| \leq \sum_{n=0}^{\infty} |a_nx^n| \leq \sum_{n=0}^{\infty} 2|x|^n$ for $|x|<1$ .
This shows that $\sum_{n=0}^{\infty} a_nx^n$ is absolutely convergent for $|x|<1.$
For $|x|=1$, the sequence $\{a_nx^n\}$ does not converge to $0$, since, in this case,  $|a_nx^n|=a_n.$
Hence $\sum_{n=0}^{\infty} a_nx^n$  is divergent for $|x|=1.$
A: We can calculate the radius of convergence with the Cauchy-Hadamard formula. Note that
$$2^{\frac{1}{2n+1}}>1$$
and decreasing, hence
$$\sup\limits_{n > k} a_n^{1/n}=2^{1/q}$$
Where $q$ is the smallest odd integer greater than $k$. This means that it's limit is $1$, so the radius of convergence is $1$. Now you are left to check $x=1$ and $x=-1$.
