I wish to find out the following infinite sum:

$$\lim_{n\to \infty}\sum_{k=0}^n\left(\frac{2}{9}\right)^k\sin\left[\frac{2\pi}{3(2^k)}\right]$$

I can sum up a GP or an AGP well, and know telescoping series, how can I find this infinite sum? Any help would be appreciated. Thanks in advance!

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    $\begingroup$ Do you have any reason to think the sum can be expressed in closed form? $\endgroup$ – Robert Israel Mar 22 '19 at 15:41
  • $\begingroup$ What is the origin of the problem? $\endgroup$ – user Mar 22 '19 at 15:42
  • $\begingroup$ @RobertIsrael I didnt get you.... $\endgroup$ – saisanjeev Mar 24 '19 at 9:36

I don't think that there is a closed form. When $k$ is sufficiently large, the argument of the sine becomes tiny and you can use a linear approximation. Hence, denoting $S$ the infinite sum (which converges),

$$\sum_{k=0}^na^k\sin(2^{-k}b)\approx S_{a,b}-\sum_{k=n+1}^\infty a^k2^{-k}b=S_{a,b}-\frac{a^{n+1}b}{2^{n}(2-a)}.$$

A better approximation is obtained by using the next term in the Taylor developments, giving $$\sum_{k=0}^na^k\sin(2^{-k}b)\approx S_{a,b}-\sum_{k=n+1}^\infty a^k2^{-k}b+\frac16\sum_{k=n+1}^\infty a^k(2^{-k}b)^3 \\=S_{a,b}-\frac{a^{n+1}b}{2^{n}(2-a)}+\frac{a^{n+1}b^3}{6\cdot8^{n}(8-a)},$$

and so on.

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  • $\begingroup$ Why not use the whole Taylor series? For $|a| <2$, $$ \sum_{n=0}^\infty a^n \sin(2^{-n} b) = \sum _{k=0}^{\infty }-2\,{\frac {{4}^{k} \left( -1 \right) ^{k}{b}^{2 \,k+1}}{ \left(a -2\cdot {4}^{k} \right) \left( 2\,k+1 \right) !}} $$ $\endgroup$ – Robert Israel Mar 22 '19 at 16:40
  • $\begingroup$ @RobertIsrael: this was implied by "and so on". ;-) $\endgroup$ – Yves Daoust Mar 22 '19 at 16:41

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