Convergent complex series

Question

Is

$$\sum\limits_{n=1}^\infty \frac{i^n}{n} $$

convergent?

Im confused as to how to solve this question, I've been trying to use ratio test but that doesn't seem to be helping.

Answer 1

Thew ratio and root tests are for series of positive terms. The series is not absolutely convergent. You can use Dirichlet's test to prove that it is (conditionally) convergent.

Answer 2

You can do this somewhat directly:

Look at $\frac{1}{4n} + \frac{i}{4n+1} + \frac{i^2}{4n+2} + \frac{i^3}{4n+3}.$ Since $i^4 = 1$, we know we can chunk up the series in this way.

So, put all of these together as one fraction: you get $\frac{(16+16i)n^2 + (16+8i)n + 3}{4n(2n+1)(4n+1)(4n+3)}$.

But this is a sum in which the terms are order $\frac{1}{n^2}$, which is certainly convergent, by eg, the comparison test applied to the absolute value of this.

This shows us that the partial sums $S_{4n}$ converge, but then the entire series must converge: Consider $|S_m - S_n|$ as two partial sums. Then, we can find the two closest $4k$ and $4j$ to $m$ and $n$, $|S_m - S_n| \leq |S_m - S_{4k}| + |S_{4k} - S_{4j}| + |S_{4j} - S_n|$. We showed that the middle term tends to zero, but then the first and last are of order $\frac{1}{n}$ and thus converge.

Moreover, this method applies for any rational multiple of $\pi$ argument.

Answer 3

The series $$A=\sum_{n\geqslant1}\frac1{4n-2}-\frac1{4n}=\sum_{n\geqslant1}\frac1{2n(4n-2)}$$ and $$B=\sum_{n\geqslant1}\frac1{4n-3}-\frac1{4n-1}=\sum_{n\geqslant1}\frac2{(4n-3)(4n-1)}$$ both converge, by comparison with the converging Riemann series $\sum\limits_{n\geqslant1}\frac1{n^2}$, and the series in the question is $-A+\mathrm i B$, hence it converges as well.

The above is a hands-on approach. One should be aware though, that for every complex number $z\ne1$ such that $|z|=1$, the series $$\sum_{n\geqslant1}\frac{z^n}n,$$ converges, and master at least one argument to show that it does, see this previous question for example.

Answer 4

The partial sum of the series to $2k$ terms is $$\frac i1-\frac12-\frac i3+\frac14+\frac i5-\frac16-\frac i7+\frac18+\cdots +(-1)^k\frac1{2k}$$ which can be rewritten as $$\Bigl(-\frac12+\frac14-\frac16+\frac18+\cdots+(-1)^k\frac1{2k}\Bigr) +i\Bigl(\frac11-\frac13+\frac15-\frac17+\cdots+(-1)^{k-1}\frac1{2k-1}\Bigr)\ .$$ Both the series are closely related to the alternating harmonic series whose limits are familiar from real calculus. The expression just given tends to $$-\frac{\ln2}{2}+i\frac\pi4$$ as $k\to\infty$. You can easily confirm that the partial sum to an odd number of terms gives the same answer, which is therefore the sum of the series.