# Convergence radius argument

I'm studying complex function theory and I ran into this argument made by my prof but I can't really wrap my head around it.

Set $f(z):=\frac{1}{7+z^2}$

Now notice

$\sum_{n=0}^\infty (\frac{-1}{7})^nz^{2n}=\sum_{n=0}^\infty (\frac{-z^2}{7})^n=\frac{1}{1+z^2/7}=\frac{7}{7+z^2}=7f(z)$

Now this converges if $|\frac{z^2}{-7}|<1$ i.e. $|z|<\sqrt{7}$.

This implies that $f(z)=\sum_{n=0}^\infty \frac{(-1)^n}{7^{n+1}}z^{2n}$ and now my prof claims that the convergence radius $R$ for this series is $R\geq \sqrt{7}$, which should follow from the calculation above. How come?

To me the convergence radius shouldn't change so it should be $R<\sqrt{7}$. Can anyone explain this?

If the radius of convergence of a power series is $R$, Then the series converges if $|z|<R$ and diverges if $|z|>R$.
The series in the question converges for $|z|<\sqrt7$. This implies that the radius of convergence is at least $\sqrt7$ (It is fact equal to $\sqrt7$.)