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Question Number 200570 by Rupesh123 last updated on 20/Nov/23

	Answered by witcher3 last updated on 20/Nov/23
	$Σ_(n≥1) ((sin(n))/n)=Im{Σ_(n≥1) (e^(in) /n)} Σ_(n=1) ^N e^(in) =((e^i (1−e^(Ni) ))/(1−e^i ))=((e^(i((1+N)/2)) sin((N/2)))/(sin((1/2)))) ∣Σ_(n≥1) e^(in) ∣≤(1/(sin((1/2)))), (1/n);n∈N^∗ decresing and cv to 0 we can applie abel theorem ⇒Σ_(n≥1) (e^(in) /n) cv ⇒Σ((sin(n))/n)=Im{Σ_(n≥1) (e^(in) /n)} =Im(−ln(1−e^i ))=−Im(ln(1−cos(1)−isin(1)) =−Imln(2sin^2 ((1/2))−2isin((1/2))cos(1/2)) =−Im (ln(2sin((1/2)))+ln(sin((1/2))−icos((1/2))) −Im(ln(e^(i((1/2)−(π/2))) )=((π−1)/2)$
	$\sum_{n ⩾ 1} \frac{\sin (n)}{n} = Im {\sum_{n ⩾ 1} \frac{e^{in}}{n}}$ $\underset{n = 1}{\sum^{N}} e^{in} = \frac{e^{i} (1 - e^{Ni})}{1 - e^{i}} = \frac{e^{i \frac{1 + N}{2}} \sin (\frac{N}{2})}{\sin (\frac{1}{2})}$ $∣ \sum_{n ⩾ 1} e^{in} ∣⩽ \frac{1}{\sin (\frac{1}{2})},$ $\frac{1}{n}; n \in N^{*} decresing and cv to 0$ $we can applie abel theorem \Rightarrow \sum_{n ⩾ 1} \frac{e^{in}}{n} cv$ $\Rightarrow Σ \frac{\sin (n)}{n} = Im {\sum_{n ⩾ 1} \frac{e^{in}}{n}}$ $= Im (- \ln (1 - e^{i})) = - Im (\ln (1 - \cos (1) - isin (1))$ $= - Imln ({2 \sin}^{2} (\frac{1}{2}) - 2 isin (\frac{1}{2}) \cos \frac{1}{2})$ $= - Im (\ln (2 \sin (\frac{1}{2})) + \ln (\sin (\frac{1}{2}) - icos (\frac{1}{2}))$ $- Im (\ln (e^{i (\frac{1}{2} - \frac{π}{2})}) = \frac{π - 1}{2}$
	Commented by Rupesh123 last updated on 20/Nov/23
	Nice one!
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