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.
['https://math.stackexchange.com/questions/2626838', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/512018/']
**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
$$
\left|\,\frac{\cos(x)}{(x+2k\pi)(x+(2k-1)\pi)}\,\right|\le\frac1{2k(2k-1)\pi^2}
$$
$$\int\_{\pi}^{+\infty}\frac{\cos x}{x}\,dx $$
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
$$ \left[\int\_{0}^{+\infty}\frac{s\,ds}{(1+s^2)^4}\right]^{1/4}\cdot\left[\int\_{0}^{+\infty}s e^{-4\pi s/3}\,ds\right]^{3/4}=\sqrt[4]{\frac{3^5}{2^{13}\,\pi^6}}\leq\frac{3}{40}. $$