Algorithm complexity using iteration method I want to find the complexity of 
$$T(n) = T\left(\frac{n}{2}\right) + n \left(\sin\left(n-\frac{π}{2}\right) +2\right)$$    by iteration method.
Assume $T(1) = 1$.
\begin{align*}
T(n) &= T\left(\frac{n}{2}\right) + n \left(\sin\left(n-\frac{π}{2}\right) +2\right)\\
&= T\left(\frac{n}{2^2}\right) + \frac{n}{2} \left(\sin\left(\frac{n}{2}-\frac{π}{2}\right) +2\right) + n \left(\sin\left(n-\frac{π}{2}\right) +2\right)\\
&= T\left(\frac{n}{2^3}\right) + \frac{n}{2^2} \left(\sin\left(\frac{n}{2^2}-\frac{π}{2}\right) +2\right) + \frac{n}{2} \left(\sin\left(\frac{n}{2}-\frac{π}{2}\right) +2\right)\\
&\mathrel{\phantom{=}}{}+ n \left(\sin\left(n-\frac{π}{2}\right) +2\right)\\
&=\cdots\\
&= T\left(\frac{n}{2^k}\right)+ n \sum_{i = 0}^{k-1} \frac{1}{2^i} \sin\left(\frac{n}{2^i} - \frac{π}{2}\right) + 2n \sum_{i = 0}^{k-1} \frac{1}{2^i}\\
&= T\left(\frac{n}{2^k}\right)+ n \sum_{i = 0}^{k-1} \frac{1}{2^i} \sin\left(\frac{n}{2^i} - \frac{π}{2}\right) + 2n \frac{1-(\frac{1}{2})^{k}}{1-\frac{1}{2}}.
\end{align*}
Now, how to I can simplify the second summation? 
Thanks.

The term to be simplified is a combination of $\sum 2^k\cos 2^k$ and $\sum 2^k\sin 2^k$ or collectively $\sum 2^ke^{i2^k}$.
As far as I know, there is no closed form expression for such a "super-geometric" summation.
Anyway, using that $|e^{it}|\le1$, the absolute value of the sum is bounded by $2n$.
