# Finding Sum of Series with $U$-substitution

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 $$2^{1-n}$$?

• You've got it. Simply substitute $x$ with $1$ and $2$ to get $2^{-n}$ and $2^{1-n}$ as limits for $u$ Commented Sep 22, 2020 at 1:26

## 1 Answer

$$S=\sum\limits_{n=1}^\infty \frac1{2^n} \int\limits_1^2 \sin \left(\frac{\pi x}{2^n}\right) dx=\sum\limits_{n=1}^\infty \frac1{2^n} \Big[-\frac{2^n}{\pi}\cos\left(\frac{\pi x}{2^n}\right)\Big]_1^2=\sum\limits_{n=1}^\infty \frac1{\pi}\left(\cos( \frac{\pi}{2^n})-\cos( \frac{2\pi}{2^n})\right)$$

The last sum consists of telescopic series so:

$$S=\dfrac{1}{\pi}\lim_{n\rightarrow \infty}\big(- cos(\pi)+cos(\frac{\pi}{2^n}\Big)=\dfrac{2}{\pi}$$