What is the limit of the nth root of a sequenc which diverges?

Question

The question under power serieses of complex numbrs. The main question is: Prove that if a given sequence converges conditionally. Then, the converging radius of The power series of that sequence equals 1.

Answer 1

We have a sequence $(a_n)_{n \ge 0}$ in $ \mathbb C$ such that $ \sum_{n=0}^{\infty}a_n$ is convergent and $ \sum_{n=0}^{\infty}|a_n|$ is divergent.

You have to show that for the radius of convergence $R$ of the power series $ \sum_{n=0}^{\infty}a_nz^n$ we have $R=1.$

1. Assume that $R>1$, then $ \sum_{n=0}^{\infty}|a_n| \cdot 1^n= \sum_{n=0}^{\infty}|a_n|$ is convergent, a contradiction.

2. Assume that $R<1$, then $ \sum_{n=0}^{\infty}a_n \cdot 1^n= \sum_{n=0}^{\infty}a_n$ is divergent, a contradiction.

Conclusion: $R=1.$