Checking for convergence of $\sum_{n=1}^\infty \frac{(-i)^n}{n}$

Question

Given the following sum, with $i$ as the imaginary unit: $$\sum_{n=1}^\infty \frac{(-i)^n}{n}$$ The sum should be checked for convergence.

I tried using the ratio test, but that failed. I think the alternating series test might be the way to go. However, the alternating series requires two things from a sum $\sum_{n=1}^\infty a_n$ for convergence, where $a_n=(-1)^nb_n$:

• $\lim_{n \to \infty} b_n = 0$
• $b_n$ is a decreasing sequence

The first requirement is met, however I don't know how to prove the second requirement, since I am unsure about the alternating thing for the imaginary unit $i$ (I know that $(-i)^n$ switches between the numbers $-i, -1, i$ and $1$, but don't know how to apply this to the alternating series test). Would really appreciate your help!

Answer 1

Just apply Dirichlet's test and use the fact that, for each natural $N$,$$\left\lvert\sum_{n=1}^Ni^n\right\rvert=\left\lvert\frac{i-i^{N+1}}{1-i}\right\rvert\leqslant\sqrt2.$$

Answer 2

For $z_{n}=a_{n}+ib_{n}$,The series $\sum_{n=1}^{\infty}z_{n}$ converges if and only if $\sum_{n=1}^{\infty}a_{n}$ and $\sum_{n=1}^{\infty}b_{n}$ converges.

$$\frac{(-i)^{n}}{n}=\frac{a_{n}}{n}+i\frac{b_{n}}{n}$$

Where $a_{n}=\begin{cases}-1\,,n=4k+2\\1,\,n=4k\\0 n=4k+1,4k+3\end{cases}$

$b_{n}=\begin{cases}-1\,,n=4k+1\\1,\,n=4k+3\\0 ,\,n=4k,4k+2\end{cases}$

So $$\sum_{n=1}^{\infty}\frac{a_{n}}{n}=\sum_{r=1}^{\infty}\frac{(-1)^{r}}{2r}$$ which converges by Leibniz Test. And does so to $-\frac{\ln(2)}{2}$

$$\sum_{n=1}^{\infty}\frac{b_{n}}{n}=\sum_{r=1}^{\infty}\frac{(-1)^{r}}{2r-1}$$ Which again converges by Leibniz Test. and does so to $-\frac{\pi}{4}$.

So as both $\sum \frac{a_{n}}{n}$ and $\sum \frac{b_{n}}{n}$ converges. You have convergence of $\sum \frac{z_{n}}{n}=\sum\frac{(-i)^{n}}{n}$.