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?
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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. |
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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? |
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