Prove that $\sum^{\infty}_{n=1}\frac{1}{n}\left(\frac{1+i}{\sqrt2}\right)^n$ converges but does not absolutely converge. Prove that $\sum^{\infty}_{n=1}\frac{1}{n}\left(\frac{1+i}{\sqrt2}\right)^n$ converges but does not absolutely converge.
My approach so far was to notice that $\left(\frac{1+i}{\sqrt2}\right)=e^{in\frac{\pi}{4}}$, so $\sum^{\infty}_{n=1}\frac{1}{n}\left(\frac{1+i}{\sqrt2}\right)^n=\sum^{\infty}_{n=1}\frac{1}{n}e^{in\frac{\pi}{4}}$.
I've also noticed that $\sum^{k+7}_{n=k}e^{i\frac{\pi}{4}}=0$ for any $k\geq1$, but I'm not sure how to proceed (maybe using the alternating series test in some way?)
Thanks.
My Solution
We can split the sum into even- and odd-indexed sums:
$\sum^{\infty}_{n=1}\frac{1}{n}\left(\frac{1+i}{\sqrt2}\right)^n=\sum^{\infty}_{k=1}\frac{1}{2k}\left(\frac{1+i}{\sqrt2}\right)^{2k}+\sum^{\infty}_{k=1}\frac{1}{2k+1}\left(\frac{1+i}{\sqrt2}\right)^{2k+1}$
$\star$But $\sum^{\infty}_{k=1}\frac{1}{2k}\left(\frac{1+i}{\sqrt2}\right)^{2k}=\sum^{\infty}_{k=1}\frac{1}{2k}\left(\frac{2i}{2}\right)^{k}=\sum^{\infty}_{k=1}\frac{1}{2k}i^{k}$, which converges by the alternating series test.
$\star$Also, $\sum^{\infty}_{k=1}\frac{1}{2k+1}\left(\frac{1+i}{\sqrt2}\right)^{2k+1}=\sum^{\infty}_{k=1}\frac{1}{2k+1}\left(\frac{1+i}{\sqrt2}\right)^{2k}\left(\frac{1+i}{\sqrt2}\right)=\sum^{\infty}_{k=1}\frac{1}{2k+1}i^k\left(\frac{1+i}{\sqrt2}\right)$, which converges by the alternating series test.
Hence, $\sum^{\infty}_{n=1}\frac{1}{n}\left(\frac{1+i}{\sqrt2}\right)^n$ converges.
However, $\left|\frac{1+i}{\sqrt2}\right|^n=1$ for all $n$, so
$\sum^{\infty}_{n=1}\frac{1}{n}\left|\frac{1+i}{\sqrt2}\right|^n=\sum^{\infty}_{n=1}\frac{1}{n}$, which is a divergent series. So, $\sum^{\infty}_{n=1}\frac{1}{n}\left(\frac{1+i}{\sqrt2}\right)^n$ does not absolutely converge.
I'm uncertain if the steps labeled $\star$ are correct.
 A: Break the sum up into even- and odd-indexed sums. The terms from the odd-indexed sums lie on the line parallel to $1+i$, and the alternating series test can be used to show that this subseries converges. Do the same with the even-indexed terms along the line parallel to $-1+i$.
A: HINT: $$\left(\dfrac{1+i}{\sqrt2}\right)^n=e^{\dfrac{n\pi i}{4}}$$ and $$\left|\dfrac{1+i}{\sqrt2}\right|^n=1.$$ Also $\sum_{n\in\mathbb{N}}\dfrac1{n}$ is divergent. 
Let $$S_{x,m}=\sum_{n=1}^m\dfrac{e^{\left(\dfrac{nx\pi i}{4}\right)}}{n},$$ then compute$\dfrac{dS_{x,m}}{dx}$ and using that you would be able to calculate $$\sum_{n\in\mathbb{N}}\dfrac{e^{\left(\dfrac{n\pi i}{4}\right)}}{n}.$$
A: Hint. First observe that sequence $A_n=\sum_{k=1}^n w^n$, where $w=\exp(i\pi/4)$ is bounded.
Then, write your sequence as
$$
\sum_{k=1}^n\frac{w^n}{n}=\sum_{k=1}^n \frac{A_n-A_{n-1}}{n}=\sum_{k=1}^n \frac{A_n}{n}-\sum_{k=1}^n \frac{A_{n-1}}{n}=\sum_{k=1}^n \frac{A_n}{n}-\sum_{k=0}^{n-1} \frac{A_{n}}{n+1}=\frac{A_n}{n}+\sum_{k=1}^{n-1}\frac{A_n}{n(n+1)}.
$$
Now both terms on the right hand side converge, as $n\to\infty$.
