Radius of convergence of unknown function

Question

I am a bit stuck on this, I think it is kind of easy, but I feel dumb because I don't see it.
Also I think something is wrong in the question.

Let $\{a_n\}_n \subseteq \mathbb{C}$ such that $\underset{n}{lim}\sqrt[n]{a_n}=\frac{1}{5}$ and $\sum_{n=0}^{\infty} a_n = i$. Calculate the radius of convergence of the series $\underset{n=0}{\overset{\infty}{\sum}} a_n(z-1)^{2n}$ and the value of

$$ \int_{C(0,3)}\frac{f(z)}{z-2} dz $$ with $f(z)=\underset{n=0}{\overset{\infty}{\sum}}a_n(z-1)^{2n}$.

I have tried with Cauchy's integral formula, but there are some things that don't fit.

Answer 1

From $\underset{n}{lim}\sqrt[n]{a_n}=\frac{1}{5}$ is straightforward that the radius of convergence is $R=5$. Now by the Cauchy's integral formula, since $2\in \operatorname{int}C(0,3)$, $$\int_{C(0,3)}\frac{f(z)}{z-2} dz=2\pi i f(2)=2\pi i\sum_{n=0}^\infty a_n(2-1)^{2n}=2\pi i\sum_{n=0}^\infty a_n=2\pi i i=-2\pi.$$