Does $\sum_{n=0}^\infty\frac{\sin(2n+1)}{2n+1}=0$?

Question

I've come to a bit of a sticking point in my answer to problem 14A given here

http://www.maths.cam.ac.uk/undergrad/pastpapers/2011/ib/List_IB.pdf

(note that this is a past paper that I am trying for myself and not homework)

Anyway, things would maybe be ok if I could easily tell that

$\sum_{n=0}^\infty\frac{\sin(2n+1)}{2n+1}=0$

Is this the case? If not what does the sum equal? Is it even convergent?

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Also, if anyone has any ideas about the end of the question I posted, where it asks me to comment on my answer, that would be much appreciated.

Answer 1

The correct answer is: $\sum_{n=0}^\infty\frac{\sin(2n+1)}{2n+1}=\frac{\pi}{4}$

You can see for example here: Prudnikov, A. P.; Brychkov, Yu. A.; and Marichev, O. I. Integrals and Series, Vol. 1: Elementary Functions. New York: Gordon and Breach, 1986. Or use Wolfram Mathematica. :)

Answer 2

Using the result at this link and the references within, you can show that $$ \sum_{n=0}^\infty\frac{\sin(2n+1)}{2n+1}={\pi\over4}. \tag1$$

To be more specific, use equation (17) on page 4 of Surprising Sinc Sums and Integrals. Plug in $N=0$ and $a_0=1$, then $N=0$ and $a_0=2$, subtract and divide by 2 to get the result (1).