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.
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You can first integrate by parts for $X > \pi$

> 
> $$
> \int\_{\pi}^{X}\frac{\cos\left(x\right)}{x}\text{d}x=\left[\frac{\sin\left(x\right)}{x}\right]^{X}\_{\pi}+\int\_{\pi}^{X}\frac{\sin\left(x\right)}{x^2}\text{d}x
> $$
> 
> 
> 

Then you can apply your inequality
$$
\left|\frac{\sin\left(X\right)}{X}\right| \leq \frac{1}{X} \underset{X \rightarrow +\infty}{\rightarrow}0
$$
and then
$$
\left|\frac{\sin\left(x\right)}{x^2}\right| \leq \frac{1}{x^2}
$$
which is integrable on $\left[\pi, +\infty\right[$. Letting $X \rightarrow +\infty$ gives you the convergence of the first integral because the three terms you find that one is constant, one tends to a constant and the last tends to $0$.
**Hint:**
$$
\begin{align}
\int\_\pi^\infty\frac{\cos(x)}{x}\,\mathrm{d}x
&=\sum\_{k=1}^\infty\int\_{(2k-1)\pi}^{(2k+1)\pi}\frac{\cos(x)}{x}\,\mathrm{d}x\\
&=\sum\_{k=1}^\infty\int\_{-\pi}^\pi\frac{\cos(x)}{x+2k\pi}\,\mathrm{d}x\\
&=\sum\_{k=1}^\infty\int\_0^\pi\left[\frac{\cos(x)}{x+2k\pi}-\frac{\cos(x)}{x+(2k-1)\pi}\right]\mathrm{d}x\\
&=-\pi\sum\_{k=1}^\infty\int\_0^\pi\frac{\cos(x)}{(x+2k\pi)(x+(2k-1)\pi)}\,\mathrm{d}x\\
\end{align}
$$
and
$$
\left|\,\frac{\cos(x)}{(x+2k\pi)(x+(2k-1)\pi)}\,\right|\le\frac1{2k(2k-1)\pi^2}
$$