Complex Convergence of $\sum\limits_{n=0}^{\infty} i^n z^n$

Question

$$\sum\limits_{n=0}^{\infty} i^n z^n$$

I am new to complex analysis, but I assume we want to apply the ratio test to form:

$$\lim_{n \to \infty} \left|\frac{i^{n+1} \ z^{n+1}}{i^n \ z^n}\right| = \lim_{n \to \infty} |i z| < 1$$

Since $|i| = 1$, we are left with the disk of convergence being $|z| < 1$. Would I also need to formally calculate the series's convergence or divergence at $z = 1, -1$?

Thanks.

Answer 1

Notice that $i=e^{i\frac{\pi}{2}}$. For $z=e^{i\theta}$, set $\theta'=\theta+\frac{]pi}{2}$. One gets that

$$ \begin{align} \sum^N_{n=0}i^nz^n=\sum^N_{n=0}e^{in(\tfrac{\pi}{2}+\theta)}=\sum^N_{n=0}e^{in\theta'}= \frac{1-e^{i(N+1)\theta'}}{1-e^{i\theta'}}\tag{1}\label{one} \end{align} $$

Recall that a if (complex) series $\sum_na_n$ converges converges, then $a_n\xrightarrow{n\rightarrow\infty}0$

In $\eqref{one}$, the $n$-th term of the summation is $a_n=e^{in\theta'}$. Since $|a_n|=1$, the series $\sum_ne^{in\theta'}$ fails to converge. This happens for any $\theta$, so the series $\sum_ni^nz^n$ diverges in $\{z:|z|=1\}$.

---

This is beyond the scope of the OP but it is worth mentioning that the the behavior of the partial sums in $\eqref{one}$ can be analyzed by looking at the behavior of the left-hand-side $$ \begin{align} S_N(\theta'):=\frac{1-e^{i(N+1)\theta'}}{1-e^{i\theta'}}\tag{2}\label{two} \end{align} $$

If $\theta'$ is rational, say $\theta'=\frac{p}{q}$, $gcd(p,q)=1$, then $S_N(\theta')$ takes $q$ different values as $N$ ranges over $\mathbb{N}$, and $S_N(\theta')$ moves trough its values periodically.

A much more interesting behavior occurs when $\theta'$ is irrational. For then, the values of $e^{i(N+1)\theta'}$ form a dense set in the unit circle $\{z:|z|=1\}$ and so, the values of the $S_N(\theta')$ form a dense set in a circle.

Answer 2

You have the right idea but you're not in $\mathbb{R}$ any more: you need to check convergence at every point on the boundary, i.e. $z=e^{i\theta}, 0\le \theta < 2\pi$. Fortunately, this is relatively easy: the terms fail the Test for Divergence (also known as the Term Test, which says if $\lim_{n\to\infty}a_n\ne 0, \sum a_n$ diverges), so the series diverges when $|z|=1$.

Answer 3

The geometric series $$\sum_{n=0}^\infty w^n$$ is known to converge for $|w|<1$ and diverge for all $|w|\ge 1.$

Let $w=iz.$