How Do You Rewrite a Power Series for Radius of Convergence?

[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?

Thanks in advance

 

[DonAntonio]

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?

Thanks in advance

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

 

[HallsofIvy]

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?

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.

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?

Thanks in advance