Summation of a trigonometric series - $\frac{\sin n}{n}$ Question:
Find the sum of the series:
$$\lim_{n \to \infty}\frac{\sin1}{1}+\frac{\sin2}{2}+\frac{\sin3}{3}+...+\frac{\sin n}{n}$$
I have no clue how to find this. Obviously I can see the sum will be convergent as the denominator gets increasingly bigger while the numerator is bound between $1$ and $-1$.
 A: Alternative Hint: consider the Fourier series of the $2\pi$-periodic function that equals $\frac{\pi-x}{2}$ over $(0,2\pi)$.
Yet another way is to use the limit case of the Euler-Maclaurin summation formula: since $\text{sinc}(x)$ is a very well-behaved even analytic function, we simply have:
$$ \sum_{n\geq 1}\frac{\sin n}{n}=\int_{0}^{+\infty}\frac{\sin x}{x}\,dx-\frac{1}{2}.$$
A: Here's a step-by-step breakdown using the fact that $\sin x = \text{Im}(e^{ix})$  
$$\lim_{n \to \infty} \sum_{k=1}^n \frac{\sin k}{k}$$
$$=\lim_{n \to \infty} \sum_{k=1}^n \frac{\text{Im}(e^{ik})}{k}$$
$$=\text{Im}\left(\sum_{k=1}^\infty \frac{e^{ik}}{k}\right) =\text{Im}\left(\sum_{k=1}^\infty \frac{(e^i)^k}{k}\right)$$
Now recall that the Taylor Series for $-\log(1-x)$ is $\sum_{k=1}^\infty \frac{x^k}{k}$
$$=\text{Im}(-\log(1-e^i))$$
$$=\text{Im}(\text{Arg}(1-e^i))$$
We now express $1-e^i$ as $1-\cos(1)-i\sin(1)$. Noting that $1-\cos(x) > 0 \;(\forall x \in \mathbb R)$ we can simply let $\text{Arg}(1-e^i) = \arctan\left(\frac{\sin(1)}{1-\cos(1)}\right) = \arctan\left(\cot(\frac{1}{2})\right)$
$$=\color{red}{\frac{1}{2}(\pi - 1)}$$
A: Hint: $\sin n$ is the imaginary part of $e^{i n}$.
A: It's a simple application of Abel-Plana formula
\begin{align}
\sum_{n = 1}^{\infty}{\sin\left(n\right) \over n} & =
-1 + \sum_{n = 0}^{\infty}{\sin\left(n\right) \over n}
\\[3mm] &=
-1 + \left[\int_{0}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x +
{1 \over 2}\,\lim_{x \to 0}{\sin\left(x\right) \over x} +
{\rm i}\int_{0}^{\infty}
{\sin\left({\rm i}n\right)/\left({\rm i}n\right) -
 \sin\left(-{\rm i}n\right)/\left(-{\rm i}n\right)\over {\rm e}^{2\pi t} - 1}\,{\rm d}t\right]
\\[3mm] &= -1 + {\pi \over 2} + {1 \over 2} + 0 =
\color{#f00}{{1 \over 2}\left(\pi - 1\right)}
\end{align}
