Determine the radius of convergence of $$\sum_{n=0}^{\infty}{\bigg(\frac{z+1}{3-i}\bigg)^{n^2}}.$$


Is this $$\sum_{n=0}^{\infty}{\bigg(\frac{z+1}{3-i}\bigg)^{n^2}}$$ even a power series? After all, we have defined power series to take the form $$\sum_{n=0}^{\infty}{{a_n}(z - c)^n}$$ where $\{a_n\}$ is a complex sequence and $c \in \mathbb{C}$ is called the center of the series.

Now, if I consider $$\sum_{n=0}^{\infty}{\bigg(\frac{z+1}{3-i}\bigg)^{n^2}}$$ to be a power series, then setting $$b_n := \bigg(\frac{z+1}{3-i}\bigg)^{n^2}$$ I get $$\left|\frac{b_{n+1}}{b_n}\right|=\left|\frac{(z+1)^{(n+1)^2}}{(3-i)^{(n+1)^2}}\right|\cdot\left|\frac{(3-i)^{n^2}}{(z+1)^{n^2}}\right|=\left|\frac{(z+1)^{2n+1}}{(3-i)^{2n+1}}\right|=\left|\frac{(z+1)^{2n}}{(3-i)^{2n}}\right|\cdot\left|\frac{z+1}{3-i}\right|.$$

So this is not the correct approach.

Now, if I use the Cauchy-Hadamard Formula for the radius of convergence $R$, I get $$\frac{1}{R} = \lim_{n \to \infty}\sup\left|b_n\right|^{1/n} = \lim_{n \to \infty}\sup\left|\bigg(\frac{z+1}{3-i}\bigg)^{n^2}\right|^{1/n} = \lim_{n \to \infty}\sup\left|\bigg(\frac{z+1}{3-i}\bigg)^n\right| = \infty.$$

Consequently, $R = 0$.

In other words, $$\sum_{n=0}^{\infty}{\bigg(\frac{z+1}{3-i}\bigg)^{n^2}}$$ converges absolutely only at $z = -1$, so that $$\text{Radius of convergence } = 0$$ and $$\text{Domain of convergence } = \left\{-1\right\}.$$


Is my attempt at determining the radius of convergence correct? If not, where is the error and how can it be mended?

  • 1
    $\begingroup$ For a less long-winded approach note that $\sum_n w^{n^2}$ converges if $|w|<1$ and diverges if $|w|>1$. $\endgroup$ – Angina Seng Sep 3 '17 at 14:34
  • $\begingroup$ Thank you for the hint, @LordSharktheUnknown! Posting an answer in a bit, I hope it is correct! =) $\endgroup$ – Archimedes Plutonium Sep 3 '17 at 14:37

Taking cue from the hint in the comments, $$\sum_{n=0}^{\infty}{\bigg(\frac{z+1}{3-i}\bigg)^{n^2}}$$ converges if $$\left|\frac{z+1}{3-i}\right| < 1,$$ which implies that $$\left|z+1\right| < \left|3-i\right| = \sqrt{10}.$$ Hence, the radius of convergence is $$R = \sqrt{10}.$$


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