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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?

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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.

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    $\begingroup$ That would have been a good response for me to use. Really can't believe I didn't see that. $\endgroup$ – vlovero Jun 13 at 2:08
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    $\begingroup$ @vlovero Your solution is fine as long as you also notice that it doesn't converge for $|x|\geq 1$ for the reasons I mentioned. Also, a small typo in yours is that it should be $|a_nx^n| \leq 2|x|^n$. Cheers. $\endgroup$ – Dzoooks Jun 13 at 2:10
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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 $x=-1$.

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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 $|x|=1.$

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