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


*

*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.
A: As we know $\dfrac{x}{2}\leqslant\sin x\leqslant x$ for $0\leqslant x\leqslant\dfrac{\pi}{2}$, so
$$\frac12=\sum_{n=1}^{\infty}\frac{1}{2^{n+1}}\leqslant\sum_{n=1}^{\infty}\sin\frac{1}{2^n}\leqslant\sum_{n=1}^{\infty}\frac{1}{2^n}=1$$
