0
$\begingroup$

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!

| cite | improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.