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

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!

• Consider real and imaginary part separately. Nov 13, 2019 at 21:19
• You can show that $S(4N)$ converge though since it has $1/n^2$ terms. Then $S(4N+1),S(4N+2),S(4N+3)$ all converge to the same limit. It is true we cannot rearrange infinite series, but we can of partial sums.
– zwim
Nov 13, 2019 at 22:11
• It's not necessary to consider real and imaginar parts separatele. Dirichlet's test works for series.of complex numbers. Nov 13, 2019 at 22:49

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

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 $$\sum \frac{z_{n}}{n}=\sum\frac{(-i)^{n}}{n}$$.