Find the sum of a series.

It's easy to check that the sum $$\sum_{n = 1}^{\infty}\sin{\frac{1}{2^n}}$$ is convergent. Can this sum be calculate precisely?

-
Functional analysis???? – András Bátkai Apr 1 '11 at 14:26
I took the liberty of retaging it as series. – Olivier Bégassat Apr 1 '11 at 15:07
Do you want to express the sum of that series in terms of which known kind of functions/operations/constructions? – Qfwfq Apr 1 '11 at 15:23
Calculated precisely means what? Is one allowed to use other mathematical constants; and if so, then what is gained? – Yemon Choi Apr 1 '11 at 22:52
Without special functions, I am afraid we can not give a direct result on this problem – yaoxiao Apr 10 '11 at 8:44

You can rewrite the series as $$\sum_{n=1}^\infty (-1)^{n+1}{1\over (2^{2n-1}-1)(2n-1)!}.$$ To do this, simply expand each term using the sine series and exchange summations. It is not a closed form, but it converges much more rapidly than the original series.
I have tried this problem by considering a more general question. Denote function $$\zeta(x) = \sum_{n = 1}^{\infty}\sin{x^n}, ~ 0 \leq x < 1.$$ It's obvious that $\zeta(0) = 0$. One can use this this sum to get various differential equations $\zeta$ satisfied. Then once $\zeta$ can be solved explicitly, the above sum is a especial case $x = \frac{1}{2}$. But I have not find a solution yet. Is this approach possible?