Find the sum of the series $$\sum_{n=1}^\infty \frac1{2^n} \int_1^2 \sin \left(\frac{\pi x}{2^n}\right) dx.$$
HINT: Simplify the $n$-th term of the series by making the substitution $u = x/2^n$ in the corresponding integral. Then consider the sequence of partial sums of the series.
I am having trouble figuring out the limits of the definite integral after $u$-substitution. I got $u=x/(2^n)$ where $u(1)=1/(2^n)=2^{-n}$ and $u(2)=2/(2^n)=2^{1-n}$.
Do I then use $2^n$ and $2^{1-n}$ as the limits of the integral or $2^{-n}$ and $2^{1-n}$?
$S=\sum\limits_{n=1}^\infty \frac1{2^n} \int\limits_1^2 \sin \left(\frac{\pi x}{2^n}\right) dx=\sum\limits_{n=1}^\infty \frac1{2^n} \Big[-\frac{2^n}{\pi}\cos\left(\frac{\pi x}{2^n}\right)\Big]_1^2=\sum\limits_{n=1}^\infty \frac1{\pi}\left(\cos( \frac{\pi}{2^n})-\cos( \frac{2\pi}{2^n})\right)$
The last sum consists of telescopic series so:
$S=\dfrac{1}{\pi}\lim_{n\rightarrow \infty}\big(- cos(\pi)+cos(\frac{\pi}{2^n}\Big)=\dfrac{2}{\pi}$