Radius of convergence of $ \sum_{k=1}^{\infty}(\frac{z^4}{4} - \frac{\pi}{7})^k$ I have the complex power series $ \sum_{k=1}^{\infty}(\frac{z^4}{4} - \frac{\pi}{7})^k$. 
Through algebraic manipulation I obtain $ \sum_{k=1}^{\infty}(\frac{1}{4})^k(z^4 - \frac{4}{7}\pi)^k$. I now argue that this is a power series around $\frac{4}{7}\pi$ with radius of convergence R = 4, using the euler root test, asserting all the while that the power 4 in the argument z doesn't affect those two quantities. However, I'm not sure how to prove or even heuristically show that last bit. In fact, I'm not even sure I'm right in asserting that the power doesn't matter, it's really more of a hunch. Does anyone know how to handle this scenario? 
 A: 
Hint: This is a geometric series
  \begin{align*}
\sum_{k=1}^\infty\left(\frac{z^4}{4}-\frac{\pi}{7}\right)^k
=\frac{1}{1-\left(\frac{z^4}{\pi}-\frac{\pi}{7}\right)}-1
\end{align*}
converging for $\left|\frac{z^4}{4}-\frac{\pi}{7}\right|<1$

A: The series in this problem is not a power series, so the concept of "radius of convergence" doesn't really apply. Because of basic results for geometric series,  the series convergeges iff $|z^4/4 - \pi/7|<1.$ The set of such $z$ is not a disc, so what does "radius" mean? This set actually looks kind of complicated.
A: Put $t=z^4$ so the series is $\sum_{k=0}^\infty \dfrac{1}{4^k}(t-t_0)^k$ where $t_0=\dfrac{4\pi}{7}$
Then the radius of convergence $R=\lim \sup \dfrac{a_k}{a_{k+1}}=4$ where $a_k=\dfrac{1}{4^k}$
Hence the series converges $\forall t$ such that $|t|<4\implies |z|^4<4\implies |z|<\sqrt 2$
Note: the radius of convergence of a power series remains unaltered if you change the point about which the series is given.So here $t_0$ has no role to play.It is needed only when you need the circle of convergence which is not needed here
So the radius of convergence is $\sqrt 2$
