TITLE: Value of $\sum\sin\frac{1}{2^i}$
QUESTION [5 upvotes]: My friend asks me a question what value is the infinite series
$$\sum\sin\frac{1}{2^i}$$
It is obviously a convergence series, however I have no idea to compute the value. So is there a value of it? Thank you.

REPLY [4 votes]: Since
$$\sin x = \sum_{j=0}^{+\infty}\frac{(-1)^j}{(2j+1)!}x^{2j+1}$$
we have that:
$$\sum_{i\geq 0}\sin\frac{1}{2^i}=\sum_{i\geq 0}\sum_{j\geq 0}\frac{(-1)^j}{(2j+1)!}\cdot\frac{1}{2^{i(2j+1)}}=\sum_{j\geq 0}\frac{(-1)^j}{(2j+1)!(1-2^{-(2j+1)})}=1.81792872\ldots$$
where the last series is very fast converging. However, I would not bet on "nicer" closed forms.

REPLY [2 votes]: One possible way is to start with the expansion of $\sin(x)$ for small values of $x$ $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}-\frac{x^{11}}{399
   16800}+O\left(x^{13}\right)$$ and replace $x$ by $\frac{1}{2^i}$. So, $$\sin(\frac{1}{2^i})=2^{-i}-\frac{2^{-3 i-1}}{3} +\frac{2^{-5 i-3}}{15} -\frac{2^{-7
   i-4}}{315} +\frac{2^{-9 i-7}}{2835}-\frac{2^{-11 i-8}}{155925}+\cdots$$ and now the summation corresponds to the sum of geometric series. Limiting to the above terms, we arrive to $$\sum_{i=0}^{\infty}\sin\frac{1}{2^i}=\frac{1167338371295774}{642125490254325}\approx 1.817928721$$
If we use $n$ terms for the Taylor expansion, we can find $$S_1=2$$ $$S_2=\frac{38}{21}\approx 1.809523810$$ $$S_3=\frac{5918}{3255}\approx 1.818125960$$ $$S_4=\frac{450902}{248031}\approx 1.817925985$$ $$S_5=\frac{10368507238}{5703472845}\approx 1.817928746$$ $$S_6=\frac{1167338371295774}{642125490254325}\approx 1.817928721$$ $$S_7=\frac{372905075405009059958}{205126345736253866925}\approx 1.817928721$$