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

Determine the radius of convergence r of the power series:

    \[ \sum_{n=0}^{\infty} \frac{z^n}{(n+1)2^n}. \]

Test for convergence at the boundary points if r is finite.

Let a_n = \frac{z^n}{(n+1)2^n}. Then we have

    \begin{align*}  \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{z^{n+1}}{(n+2)2^{n+1}} \right) \left( \frac{(n+1)2^n}{z^n} \right) \\[9pt]  &= \lim_{n \to \infty} \frac{(n+1)z}{2(n+2)} \\  &= \frac{z}{2}. \end{align*}

This implies r = 2. So, the series converges for all |z| < 2. Furthermore, the series is convergent at every boundary point |z| = 2 except z = 2 by Dirichlet’s test (Theorem 10.17). Hence, the series converges for all |z| \leq 2 with z \neq 2.


  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.

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