Convergence of $\sum_{n=1}^{\infty} \frac{{i}^{3^n}}{n}$

Question

I’m given a problem for homework and I want to as how to solve it:

does the following sum converge? $$\sum_{n=1}^{\infty} \frac{{i}^{3^n}}{n}$$

It converges since ${i}^{3^n} $ is $-i, \space i, \space -i, \space i, \space -i, \space ...$ and it is $i (-1)^n$. So our given sum is equivalent to the sum $$i \sum_{n=1}^{\infty} \frac{ (-1)^n}{n} $$. But how can I prove without using wolfram alpha that ${i}^{3^n} \equiv i (-1)^n$ ?

Answer 1

Proof by induction. The base case $n=1$ is clear. Now suppose that the claim holds for all positive integers less than or equal to $n$. Then $$ i^{3^{n+1}}=(i^{3^{n}})^3\stackrel{\text{IH}}{=}(i(-1)^n)^3=-i(-1)^{3n}=i(-1)^{3n+1}=i(-1)^{n+1} $$ since $3n+1\equiv n+1$ modulo $2$, where $\text{IH}$ refers to induction hypothesis.

Answer 2

Note that for $n\geq 0$, $$ i^n=\begin{cases} 1&n\cong0\mod4\\ i&n\cong1\mod4 \\ -1&n\cong2\mod4 \\ -i&n\cong3\mod4\\ \end{cases} $$ and $$ 3^n\cong(-1)^n\mod 4 $$ so your sum converges since the alternating harmonic series converges.

Answer 3

First of all, i^4k+a = i^a. That is, iⁿ = i^{n mod 4}. But note that mod 4, 3 = -1. So i³ⁿ = i^(-1)n = i(i^(-1)n-1). Since (-1)ⁿ = (-1)^p where p = n mod 2, we can write this as i(i^(-1)p-1). If p = 0, then i^(-1)p-1 is 1, and if p = 1, this is -1.