# 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?

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.

• That would have been a good response for me to use. Really can't believe I didn't see that. – vlovero Jun 13 at 2:08
• @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. – Dzoooks Jun 13 at 2:10

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 $$x=-1$$.

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 $$|x|=1.$$