# Radius of convergence of unknown function

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.

• $R=5$ and the integral is $(2\pi i) f(2)=-2\pi$. May 28 at 12:28

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 $$2\in \operatorname{int}C(0,3)$$, $$\int_{C(0,3)}\frac{f(z)}{z-2} dz=2\pi i f(2)=2\pi i\sum_{n=0}^\infty a_n(2-1)^{2n}=2\pi i\sum_{n=0}^\infty a_n=2\pi i i=-2\pi.$$
• but it shouldn't be the case that $1/R=\underset{n}{lim}\sqrt[2n]{a_n}$ and so $R=\sqrt{5}$? I think this was the thing that was wrong in the setting of the question. I knew that something was off, because at the time in the exam where I got this question, some classmate said that something wasn't right. Jun 3 at 16:24