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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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Functional analysis???? –  András Bátkai Apr 1 '11 at 14:26
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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
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Without special functions, I am afraid we can not give a direct result on this problem –  yaoxiao Apr 10 '11 at 8:44
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2 Answers 2

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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Alternatively, one could succesively compute values of 1/2^m + sum(j=1 to m) sin (1/2^j) until one got tired/satisfied with the amount of effort done. Gerhard "Some Do This For Fun?" Paseman, 2011.04.01 –  Gerhard Paseman Apr 1 '11 at 17:12
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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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