What would be the radius of convergence of ∑n=0∞z3n ?

Question

queueryutisteh7n · Accepted Answer

What do you do for the cosine and sine series? There, you cannot use the Ratio Test directly because every other coefficient is equal to 0. Instead, we do the Ratio Test on the subsequence of even (resp. odd) terms. You can do the same here. We have

a_{3^{n}} = 1

for all n, and

a_{j} = 0

for j not a power of 3. Define

b_{k} = z^{3^{k}}

. Then we are trying to determine the convergence of the series

\sum b_{k}

, so using the Ratio Test we have:

lim_{n \to \infty} \frac{| b_{n + 1} |}{| b_{n} |} = lim_{n \to \infty} \frac{{| z |}^{3^{n + 1}}}{{| z |}^{3^{n}}} = lim_{n \to \infty} {| z |}^{3^{n + 1} - 3^{n}} = lim_{n \to \infty} {| z |}^{2 \times 3^{n}}

= {\begin{matrix} 0 if | z | < 1 \\ 1 if | z | = 1 \\ \infty if | z | > 1 \end{matrix}

So the radius of convergence is 1.

What would be the radius of convergence of

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