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!

  • 2
    $\begingroup$ Consider real and imaginary part separately. $\endgroup$ Nov 13, 2019 at 21:19
  • $\begingroup$ 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. $\endgroup$
    – zwim
    Nov 13, 2019 at 22:11
  • $\begingroup$ It's not necessary to consider real and imaginar parts separatele. Dirichlet's test works for series.of complex numbers. $\endgroup$ Nov 13, 2019 at 22:49

2 Answers 2


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.


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


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