Prove $i$ is a point of accumulation Let $a_{n} = 2^{-\frac {n}{2}}(1+i)^{n} \frac {1+n}{n}$ for  n$ \in \mathbb N.$
In order to prove the $i$ is a point of accumulation of the sequence ${a_{n}}$, I realize that I need to construct/find a subsequence $a_{l}$ where $l \in \mathbb N$. I am perplexed with regards to $(1+i)^n$ as it does not seem to maintain an $i$ for any such $a_{l}$. 
Am I on the right track? Any help would be greatly appreciated. 
 A: We have $a_{2m} = 2^{-m}(1+i)^{2m} \frac{1+2m}{2m}$. What can you say about $2^{-m}(1+i)^{2m}$? What can you say about $\frac{1+2m}{2m}$?
(Note that $a_{2m}$ is not a subsequence that converges to $i$, you have to further restrict the indices. However, it's a reasonable starting point.)
Also, I'm pretty sure that your subsequence won't be of the form $a_l$ with $l=kn$, as you suggest!
A: $$a_{n} = 2^{-\frac {n}{2}}(1+i)^{n} \frac {1+n}{n}=(\dfrac{1+i}{\sqrt 2})^n\dfrac{1+n}{n}=e^{i\dfrac{\pi n}{4}}\dfrac{n+1}{n}$$therefore $$a_{8n+2}=i\dfrac{8n+3}{8n+2}$$
A: $1+i=\sqrt{2}\exp(\frac{i\pi}4)\quad$ so $a_n=(1+\frac 1n)\exp(i\frac{n\pi}4)=(1+\frac 1n)c_n$
The sequence $c_n$ is cyclic
$\begin{cases}
c_{8n}=1 & c_{8n+4}=-1 \\
c_{8n+1}=\frac 1{\sqrt{2}}(1+i) & c_{8n+5}=\frac {-1}{\sqrt{2}}(1+i) \\
c_{8n+2}=i & c_{8n+6}=-i \\
c_{8n+3}=\frac {-1}{\sqrt{2}}(1-i) & c_{8n+7}=\frac 1{\sqrt{2}}(1-i) \\
\end{cases}$
So each of these values will be an accumulation point for $a_n$ given that $1+\frac 1n\to 1$ (in other words $1$ is an accumulation point of $1+\frac 1n$).
In the present case $a_{8n+2}\to i$

At first you wrote a subsequence as being $a_{kn}$, but it is more general than this. 
A subsequence is $a_{f(n)}$ where $f$ is a strictly increasing function of $n$.
With cyclic cases like this one, $f$ is linear: $f(n)=an+b$ 
But it could be $f(n)=n^2$ or $f(n)=n!$ or even something more random than this, all is required is that $f$ is $\nearrow$.
