Does $\sum_{k=1}^{\infty} \frac{1}{k}\cdot(\frac{1+i}{1-i})^k$ converge, absolutely converge?

Question

$\sum_{k=1}^{\infty} \frac{1}{k}\cdot(\frac{1+i}{1-i})^k$, $a_k=\frac{1}{k}\cdot(\frac{1+i}{1-i})^k \in\mathbb{C}$

What I figured out so far:

$a_k=\left\{\begin{array}{0} \frac{1}{k}, & k\equiv0 \text{ (mod } 4) \\ \frac{1}{k}i, & k\equiv1 \text{ (mod } 4)\\ -\frac{1}{k}, & k\equiv2 \text{ (mod } 4) \\ -\frac{1}{k}i, & k\equiv3 \text{ (mod } 4)\end{array}\right. $

I know that the series converges if the sequence of partial sums $s_n:=\sum_{k=1}^{n} a_k$ converges.

So $s_1 = -1, s_2=-\frac{3}{2}, s_3=\frac{5}{6}, s_2=\frac{13}{12}, s_2=\frac{53}{60},s_2=-\frac{7}{60}$.

Therefore I assume that $\lim_{n\to\infty} s_n$ exists.

I don't know how to handle the different cases for $a_k$ in my $s_n$.

I appreciate any hints.

Answer 1

The series can be written as $$\sum _{k=1}^{\infty } \frac{i^k}{k}=\sum _{k=1}^{\infty } \frac{i^{2 k}}{2k}+\sum _{n=0}^{\infty } \frac{i^{2 k+1}}{2k+1}$$ The first one can be rewritten as $$\sum _{k=1}^{\infty } \frac{i^{2 k}}{2k}=\sum _{k=1}^{\infty } \frac{(-1)^k}{2k}=-\frac{\log 2}{2}$$ which converges for the Leibniz test.

The second one can be rewritten as $$\sum _{n=0}^{\infty } \frac{i^{2 k+1}}{2k+1}=i\sum _{n=0}^{\infty } \frac{(-1)^k}{2k+1}=\frac{i\pi}{4}$$ Thus the series converges to $$\sum _{k=1}^{\infty } \frac{i^k}{k}=-\frac{\log (2)}{2}+\frac{i \pi }{4}$$ As detailed in the answer above, the series does not converge uniformly because the absolute values series is the harmonic series which diverges.

Answer 2

hint

Observe that $$\frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1-i)(1+i)}$$ $$=\frac{2i}{2}=i$$

and $$\left|\frac 1k \left(\frac{1+i}{1-i}\right)^k\right|=\frac 1k$$

with $$\sum \frac 1k \;\; divergent$$

Answer 3

You have $\left|\frac{1+i}{1-i}\right|=\frac{|1+i|}{|1-i|}=1$ and so, for each $k\in\Bbb N$,$$\left|\frac1k\left(\frac{1+i}{1-i}\right)^k\right|=\frac1k.$$Therefore your series doesn't converge absolutely.

But it converges, by Dirichlet's test: the sequence $\left(\frac1k\right)_{k\in\Bbb N}$ is monotonic and converges to $0$ and the partial sums of the series $\sum_{k=1}^\infty\left(\frac{1+i}{1-i}\right)^k$ are bounded, since, for each $K\in\Bbb N$,$$\left|\sum_{k=1}^K\left(\frac{1+i}{1-i}\right)^k\right|=\left|\frac{\frac{1+i}{1-i}-\left(\frac{1+i}{1-i}\right)^{K+1}}{1-\frac{1+i}{1-i}}\right|\leqslant\frac2{|1-i|}=\sqrt2.$$

Answer 4

Consider the function $f(z)=\frac 1 {1+z}$. Observe that, for $|z|<1$, we have $f(z)=\sum\limits_{n=0}^{\infty} {(-z)^n}$, integrating from $0$ to $w$ (with $|w|\leq 1$, we obtain $\log(w+1)=\int_0^w f = \sum_{n=0}^\infty \frac{-(-z)^{n+1}}{n+1} \Bigr|_0^w$. What happens when you take $w=-i$?