Find the result of $\sum_{n=1}^{\infty} \sin\left(\frac{1}{2^n}\right)$

Question

Applying ratio test, we can prove this series $\displaystyle \sum_{n=1}^{\infty} \sin\left(\frac{1}{2^n}\right)$ converges.

How can we calculate or estimate the sum?

Any help is appreciated, thank you.

Answer 1

As we know $\dfrac{x}{2}\leqslant\sin x\leqslant x$ for $0\leqslant x\leqslant\dfrac{\pi}{2}$, so

$$\frac12=\sum_{n=1}^{\infty}\frac{1}{2^{n+1}}\leqslant\sum_{n=1}^{\infty}\sin\frac{1}{2^n}\leqslant\sum_{n=1}^{\infty}\frac{1}{2^n}=1$$

Answer 2

$$\sum_{n=1}^{\infty} \sin\left(\frac{1}{2^n}\right)$$

• For $n\ge1$, $0<\sin\left(\dfrac{1}{2^n}\right)<\dfrac{1}{2^n}\implies$ the given sum is convergent (in fact, absolutely convergent)

• To have some fun with it, we can write:

$$\sum_{n=1}^{\infty} \sin\left(\frac{1}{2^n}\right)=\sum_{n=1}^{\infty} \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!2^{n(2k+1)}}$$

$$=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} \sum_{n=1}^{\infty}2^{-n(2k+1)} =\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}\frac{1}{2^{2k+1}-1}$$

This isn't an "exact" form for the sum, but is fairly quick to converge, if you're interested in computing the sum.

Answer 3

Suppose we want to compute the series with an error of at most $\epsilon>0$. We know (why?) that for $x>0$ we have $x-\frac16x^3<\sin x<x$. Hence for the tail of the series after the $N$th summand we have the bounds $$\sum_{n=N+1}^\infty\left(2^{-n}-\frac16\cdot 8^{-n} \right)<\sum_{n=N+1}^\infty \sin(2^{-n})<\sum_{n=N+1}^\infty2^{-n},$$ an using the formula for the geometric series $$2^{-N}-\frac16\cdot 7^{-N}<\sum_{n=N+1}^\infty\left(2^{-n}-\frac16\cdot 8^{-n} \right)<\sum_{n=N+1}^\infty \sin(2^{-n})<2^{-N}.$$ Thus it suffices to pick $N\ge \log_7 \frac{12}\epsilon $ (e.g., with $\epsilon =10^{-9}$, we can use $N=12$) and then use $$ 2^{-N}-\frac1{12}\cdot 7^{-N}+\sum_{n=1}^N\sin2^{-n}$$ as approximation (provided, the computational error for computing the sines is small enough).