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I need to find the radius of convergence of the following complex series: $$\sum_{n}{\sqrt{n}(4 + (-1)^n)z^n}.$$

What I did is this:

for $n$ even: $c_n = \sqrt{n}(4 + 1) = 5\sqrt{n}$

$\limsup \frac{c_{n+1}}{c_n} = 1$

and I get the same thing when $n$ is odd

since $\limsup \sqrt{c_n} \leq \limsup \frac{c_{n+1}}{c_n}$.

The radius of convergence is then $1/1 = 1$.

is this in any way correct?

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No, your answer is not fully correct.

The radius of convergence is 1 because by the Squeeze Theorem, $$1\leftarrow(3\sqrt{n})^{1/n} \leq (c_n)^{1/n}\leq (5\sqrt{n})^{1/n}\to 1$$ where $c_n:=\sqrt{n}(4 + (-1)^n)>0$. Moreover, $$\lim_{n\to +\infty}\frac{c_{2n+1}}{c_{2n}}=\frac{3}{5}\quad\mbox{and}\quad \lim_{n\to +\infty}\frac{c_{2n+2}}{c_{2n+1}}=\frac{5}{3}.$$ which implies that $$\liminf_{n\to +\infty}\frac{c_{n+1}}{c_{n}}=\frac{3}{5}\quad\mbox{and}\quad \limsup_{n\to +\infty}\frac{c_{n+1}}{c_{n}}=\frac{5}{3}.$$

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  • $\begingroup$ so the radius of convergence is actually $\frac{3}{5}$ ? $\endgroup$
    – user368063
    Commented Oct 21, 2016 at 17:25
  • $\begingroup$ please can you explain the implication of the last 2 lines? $\endgroup$
    – user368063
    Commented Oct 21, 2016 at 18:52
  • $\begingroup$ @HexaFlexagon The implication is that in this case the limit of the ratio does not exists (it is useless for finding the radius of convergence). $\endgroup$
    – Robert Z
    Commented Oct 21, 2016 at 19:47
  • $\begingroup$ thank you so much, it's clear now $\endgroup$
    – user368063
    Commented Oct 21, 2016 at 19:58

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