# A Question about Power Series

1. Oct 25, 2012

### Artusartos

If we have the series $\sum_{n=0}^{\infty} 2^{-n}x^{3n}$. We need to calculate the radius of convergence...the textbook says that, since the power of x is 3n instead of 3, we need to rewrite it in the form $\sum_{n=0}^{\infty} a^{n}x^{n}$..and we say that $a_{3k}=2^{-k}$. I'm not sure if I understand this...so are they setting n=3k? But we know that the power of 2 is -n, so if n=3k then doesn't the power of 2 need to be -3k?

2. Oct 25, 2012

### DonAntonio

Bah. Apply directly D'Alembert's (quotient or ratio) test with $\,a_n=2^{-n}x^{3n}\,$:

$$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{x^{3n+3}}{2^{n+1}} \cdot \frac{2^n}{x^{3n}}\right|=\frac{|x|^3}{2} \xrightarrow [n\to\infty] {}\frac{|x|^3}{2}<1\Longleftrightarrow |x|<\sqrt [3] 2$$

and there you have your convergence radius. There's only left to know whether either of the convergence interval's extreme

points are contained in it.

DonAntonio

3. Oct 25, 2012

### HallsofIvy

Staff Emeritus
No. They are thinking of $x^{3n}$ as $(x^3)^n$ so that they will get the radius of convergence in terms of $x^3$ rather than just x. You can then take the cube root.