# Problem dealing with $\sum \frac{\sin(n)}{n}$ and its convergence [duplicate]

$$\text{If} \ S=\displaystyle\sum_{n=1}^{\infty}\dfrac{\sin (n)}{n}, \ \text{then what is} \ 2S+1$$

I know that $\sum \frac{\sin(n)}{n}$ converges. But now what do I do?

## marked as duplicate by Thomas Andrews, JMP, Claude Leibovici calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 14 '15 at 5:23

• Hint: what is the value of the sum from $-1$ to $-\infty$? – Steven Stadnicki Nov 14 '15 at 4:32
I'm not going to prove this here, but we know that (ask another question if you don't): $$\sum_{k=1}^{\infty} \frac{x^k }{k} = -\log(1-x)$$ For any $x$ s.t. $|x|\leq 1, x\neq 1$: $$S= \text{Im} \sum_{n=1}^{\infty} \frac{e^{i n}}{n} = -\text{Im}(\log(1-e^{i} ))$$ Now with some trig: $$1-e^i = 1-\cos 1 -i \sin 1 = 2\sin \frac{1}{2}\left(\sin \frac{1}{2} - i \cos \frac{1}{2} \right) = 2\exp\left(\frac{(1-\pi)i}{2}\right) \sin \frac{1}{2}$$ Therefore, $$S= \text{Im}(\log(1-e^{i} )) = \text{Im}\left(\frac{(\pi-1)i}{2} - \log \left(2\sin \frac{1}{2}\right) \right) = \frac{\pi-1}{2}$$ $$\therefore 2S+1 =\pi$$