MHB Radius of Convergence for $\sum_{j=0}^{\infty} \frac{z^{2j}}{2^j}$

[Guest2]

Radius of convergence of $\displaystyle \sum_{j=0}^{\infty} \frac{z^{2j}}{2^j}$.

If I let $z^2 = x$ I get a series whose radius of convergence is $2$ (by the ratio test).

How do I get from this that the original series has a radius of convergence equal to $\sqrt{2}$?

 

[Greg]

$$\dfrac{z^{2j}}{2^j}=\left(\dfrac{z}{\sqrt2}\right)^{2j}$$, so the series is a geometric series which converges for all $$|z|<\sqrt2$$.

 

[Guest2]

Thank you.

Cauchy root test gives: $\displaystyle (\lim |a_n|^{1/n})^{-1} =\left(\lim \left|\frac{x^{2n}}{2^n}\right|^{1/n}\right)^{-1} = \lim \left|\frac{x^2}{2}\right| = \frac{1}{2} |x|$ and $\displaystyle \frac{1}{2} |x^2| < 1 \implies |x| < \sqrt{2}. $ So $R = \sqrt{2}$

However, my book says $R = (\lim |a_n|^{1/n})^{-1})$ (if it exists). Surely, that can't be right as I still have $x$ term in the limit?

 

[Prove It]

$$\dfrac{z^{2j}}{2^j}=\left(\dfrac{z}{\sqrt2}\right)^{2j}$$, so the series is a geometric series which converges for all $$|z|<\sqrt2$$.

Technically it would need to be written as $\displaystyle \begin{align*} \left( \frac{z^2}{2} \right) ^j \end{align*}$ to be read off as a geometric series. So it converges where the absolute value of the common ratio $\displaystyle \begin{align*} \left| \frac{z^2}{2} \right| < 1 \implies \left| z^2 \right| < 2 \implies \left| z \right| < \sqrt{2} \end{align*}$...