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If \(\displaystyle S = \sum_{n=1}^{\infty}\frac{\sin (n)}{n}.\) Then value of \(\displaystyle 2S+1 = \)

The \(2\pi\)-periodic function \(f:{\mathbb{R}}\longrightarrow{\mathbb{R}}\,;\) \[f(x)=\left\{{\begin{array}{ll}
\dfrac{1}{2}\Bigl({\pi-x+2\pi\Bigl\lfloor{\frac{x}{2\pi}}\Bigr\rfloor}\Bigr)\,,& x\in\bigl(2k\pi,2(k+1)\pi\bigr)\\
0\,, & x=2k\pi
\end{array}}\right.\,,\quad k\in{\mathbb{Z}}\,,\] is integrable on \(\mathbb{R}\) and odd and for that has Fourier representasion:
\[f(x)=\mathop{\sum}\limits_{n=1}^{+\infty}{\frac{1}{n}\sin(nx)}\,,\quad x\in\mathbb{R}\,.\] So \begin{align*}
f(1)=\frac{1}{2}({\pi-1})=S&=\mathop{\sum}\limits_{n=1}^{+\infty}{\frac{\sin{n}}{n}}\quad\Rightarrow\\
2S+1&=\pi\,.
\end{align*}

The more general result for the series

$$ \sum_{n=1}^{\infty} \frac{\sin n \theta}{n}=\frac{\pi-\theta}{2} , \quad \theta \in (0, 2\pi) $$

can also be obtained from the Fourier expansion of the $2\pi$ periodic function $ f:\mathbb{R} \longrightarrow \mathbb{R}$ such that $f(x)=\left\{\begin{matrix}
1 &, &\left | x \right |\leq \delta \\
0&, &\delta<x \leq \pi
\end{matrix}\right.$.

Now if we plug in \( \theta=1\) we get that

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

Thus $ 2S+1=2\cdot \frac{\pi-1}{2}+1=\pi$ as in Grigoris solution.