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

Using partial summation. Obviously,

$$S = \lim_{n \to \infty} \sum_{x=1}^{n} \frac{\sin(x)}{x}$$

Partial Summation:

\begin{align*} \sum_{n=1}^{N} a(n) f(n) & = \sum_{n=1}^{N} f(n) (A(n)- A(n-1)) = \sum_{n=1}^{N} A(n) f(n) - \sum_{n=1}^{N} A(n-1) f(n)\\ & = \sum_{n=1}^{N} A(n)f(n) - \sum_{n=0}^{N-1} A(n) f(n+1)\\ & = A(N)f(N) - A(0) f(1) - \sum_{n=1}^{N-1} A(n) (f(n+1)-f(n)) \end{align*}

But how do I apply it? What is $A(n)$? I am confused?

Marco Cantarini
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Lebes
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  • Look up Dirichlet test in Wikipedia – Alex Feb 21 '15 at 20:05
  • @Alex, but how will that helped (I already knew about this)? – Lebes Feb 21 '15 at 20:10
  • I'm sorry, maybe I confused something...perhaps this could help: http://math.stackexchange.com/questions/225941/proving-sum-limits-k-0n-coskx-frac12-frac-sin-frac2n12x – Alex Feb 21 '15 at 20:19

1 Answers1

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$A\left(n\right)$ in your case can be$$\sum_{n=1}^{N}\sin\left(n\right)=\sin\left(\frac{N+1}{2}\right)\sin\left(\frac{N}{2}\right)\textrm{cosec}\left(\frac{1}{2}\right)$$ an so $$f\left(n\right)=\frac{1}{n}$$ or$$\sum_{n=1}^{N}\frac{1}{n}=H_{N}$$ where $H_{N}$ is the $N$ -th armonic number, and$$f\left(n\right)=\sin\left(n\right)$$ but I think there is a more rapid way using Fourier series. In fact we have, for $0<x<2\pi$ $$\frac{\pi-x}{2}=\sum_{n\geq1}\frac{\sin\left(nx\right)}{n}.$$

Marco Cantarini
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