8
$\begingroup$

I noticed that a lot of the time, people ask whether the following sum converges:

$$\sum_{n=1}^\infty\frac{\sin(n)}n$$

Though I've never stopped to ask what it equaled. According to this other post, the sum is given as

$$\sum_{n=1}^\infty\frac{\sin(n)}n=\frac{\pi-1}2$$

The solution involves realizing $\sin(n)=\Im e^{in}$ and the Taylor expansion for the natural logarithm.

While thats great and all, how can I prove this using only real numbers?

$\endgroup$
  • 1
    $\begingroup$ Find the (real) Fourier series of the function $x \mapsto \frac{\pi- x}{2}$ on $[0,2\pi]$. $\endgroup$ – Daniel Fischer Nov 26 '16 at 15:26
  • $\begingroup$ @DanielFischer If you could do that that'd be great $\endgroup$ – Simply Beautiful Art Nov 26 '16 at 15:30
  • 3
    $\begingroup$ How about this? $\endgroup$ – Daniel Fischer Nov 26 '16 at 15:33
  • $\begingroup$ @DanielFischer Oh, thanks for the nice find. $\endgroup$ – Simply Beautiful Art Nov 26 '16 at 15:37
  • $\begingroup$ I thought Fourier transformation involve Complex numbers. $\endgroup$ – Zaid Alyafeai Sep 4 '17 at 4:35
6
$\begingroup$

As suggested in comments, lets use fourier series. =). From here we have the fourier series of $x$, valid in the range $[-\pi, \pi]$: $$ x = -2\sum_{n=1}^\infty\frac{(-1)^{n}}{n}\sin(nx) $$

If we insert: $x=\pi-1$, it will elliminate the $(-1)^n$ from the formula. $$ \sin(nx) = \sin(n\pi - n) = \sin(n\pi)\cos(n)-\cos(n\pi)\sin(n) = -(-1)^n\sin(n) $$

Then: $$ \pi-1 = 2\sum_{n=1}^\infty\frac{1}{n}\sin(n) \quad\implies\quad \sum_{n=1}^\infty\frac{\sin(n)}{n} = \frac{\pi-1}{2} $$

$\endgroup$

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.