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 $0$.
$$\int\_{\pi}^{+\infty}\frac{\cos x}{x}\,dx $$
is convergent by [Dirichlet's test](https://en.wikipedia.org/wiki/Dirichlet%27s_test) since $\left|\int\_I\cos(x)\,dx\right|\leq 2$ and $\frac{1}{x}$ decreases to zero on $x\geq \pi$.  

Accurate upper bounds can be deduced [from the Laplace transform](https://en.wikipedia.org/wiki/Laplace_transform#Evaluating_integrals_over_the_positive_real_axis) and Holder's inequality. Indeed
$$ \int\_{\pi}^{+\infty}\frac{\cos(x)}{x}\,dx = \int\_{0}^{+\infty}\frac{-\cos x}{x+\pi}\,dx = -\int\_{0}^{+\infty}\frac{s e^{-\pi s}}{1+s^2}\,ds<0 $$
but
$$ \int\_{0}^{+\infty}\frac{s e^{-\pi s}}{1+s^2}\,ds = \int\_{0}^{+\infty}\left(\frac{s^{1/4}}{1+s^2}\right)^1\cdot\left(s^{1/4} e^{-\pi s/3}\right)^3\,ds $$
is bounded by
$$ \left[\int\_{0}^{+\infty}\frac{s\,ds}{(1+s^2)^4}\right]^{1/4}\cdot\left[\int\_{0}^{+\infty}s e^{-4\pi s/3}\,ds\right]^{3/4}=\sqrt[4]{\frac{3^5}{2^{13}\,\pi^6}}\leq\frac{3}{40}. $$