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Determine the radius of convergence of ∑ (z+3)n / (n+1)2n

Determine the radius of convergence r of the power series:

    \[ \sum_{n=0}^{\infty} \frac{(z+3)^n}{(n+1)2^n}. \]

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


Let a_n = \frac{(z+3)^n}{(n+1)2^n}. Then

    \begin{align*}  \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(z+3)^{n+1}}{(n+2)2^{n+1}} \right) \left( \frac{(n+1)2^n}{(z+3)^n} \right) \\[9pt]  &= \lim_{n \to \infty} \frac{(z+3)(n+1)}{2(n+2)}. \end{align*}

This implies the series converges for |z+3| < 2 which implies r = 2. Furthermore, the series converges on the boundary except the point z = -1. Hence, the series converges for all z such that |z+3| \leq 3 and z \neq -1.

One comment

  1. Anonymous says:

    I think there is a typo in the conclusion: the series converges for all $z$ such that $|z+3| \leq 2$, not 3

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