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# Determine the radius of convergence of ∑ zn / ((n+1)2n)

Determine the radius of convergence of the power series: Test for convergence at the boundary points if is finite.

Let . Then we have This implies . So, the series converges for all . Furthermore, the series is convergent at every boundary point except by Dirichlet’s test (Theorem 10.17). Hence, the series converges for all with .

### 2 comments

1. tom says:

I don’t recall any examples using |z| other then 1 so the boundary points |z|=2, z≠2 satisfied using Dirichlet’s test is new to me. I assume using z^n = r^n(cos nθ + i sin nθ) with θ=0 is the key, separating the series into real and imaginary parts of which the real part diverges. But at θ=π would we not be using Leibniz’s instead of Dirichlet’s test to prove convergence?

• tom says:

Some more thoughts: Apostol theorem 10.19 stipulates θ not equal to multiples of π, so Dirichlet’s test can’t be used at z=-1? So then Leibniz’s test could be used for z=-1. But my opinion- Apostol didn’t really give much attention to what seems a delicate matter, that is the radius of convergence for complex power series. Perhaps it should be obvious but I would imagine an appeal to De Moivre’s theorem with the rules for convergence of complex series would have been nice.