Evaluate Σ ((sin(3n))/n) from 1 to infinity


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Question Number 7191 by Tawakalitu. last updated on 15/Aug/16

$E v a l u a t e Σ \frac{s i n (3 n)}{n} f r o m 1 t o i n f i n i t y$
	Answered by Yozzia last updated on 15/Aug/16
	$Define the function f(x)=x for 0<x<1 , period=2. For Fourier series of f having the form (1) f(x)=(a_0 /2)+Σ_(n=1) ^∞ {a_n cos((nπx)/L)+b_n sin((nπx)/L)}, 2L=2=period⇒L=1, a_0 =(1/L)∫_c ^( c+2L) f(x) dx , c∈R. ∴ For c=0, a_0 =(1/1)∫_0 ^( 2) xdx=(x^2 /2)∣_0 ^2 =2 ⇒(a_0 /2)=1. a_n =(1/L)∫_c ^(c+2L) f(x)cos((nπx)/L)dx n=1,2,3,... Let c=0. ∴ a_n =(1/1)∫_0 ^( 2) xcosnπx dx a_n =(x/(nπ))sinnπx∣_0 ^2 −∫_0 ^2 (1/(nπ))sinnπxdx a_n =(1/(n^2 π^2 ))cosnπx∣_0 ^2 =(1/(n^2 π^2 ))(cos2nπ−1)=0 b_n =(1/L)∫_c ^( c+2L) f(x)sin((nπx)/L)dx (n=1,2,3,...) Take c=0. ∴ b_n =(1/1)∫_0 ^( 2) xsinnπx dx b_n =((−xcosnπx)/(nπ))∣_0 ^2 −∫_0 ^2 ((−cosnπx)/(nπ))dx b_n =((−2)/(nπ))+[(1/(n^2 π^2 ))sinnπx]_0 ^2 b_n =((−2)/(nπ))+0=((−2)/(nπ)) (n≠0). ∴ in (1) x=1+Σ_(n=1) ^∞ ((−2sinnπx)/(nπ)) x=1−(2/π)Σ_(n=1) ^∞ ((sinnπx)/n) Let x=(3/π)∉Z (If x∈Z, x is a point of discontinuity whose output is given by ((f(x+0)+f(x−0))/2)) ∴ (3/π)=1−(2/π)Σ_(n=1) ^∞ ((sin3n)/n) ⇒Σ_(n=1) ^∞ ((sin3n)/n)=(π/2)(1−(3/π)) Σ_(n=1) ^∞ ((sin3n)/n)=((π−3)/2)$
	$D e f i n e t h e f u n c t i o n$ $f (x) = x f o r 0 < x < 1, p e r i o d = 2 .$ $F o r F o u r i e r s e r i e s o f f h a v i n g t h e f o r m$ $(1) f (x) = \frac{a_{0}}{2} + \underset{n = 1}{\sum^{\infty}} {a_{n} c o s \frac{n π x}{L} + b_{n} s i n \frac{n π x}{L}},$ $2 L = 2 = p e r i o d \Rightarrow L = 1,$ $a_{0} = \frac{1}{L} \int_{c}^{c + 2 L} f (x) d x, c \in R .$ $∴ F o r c = 0, a_{0} = \frac{1}{1} \int_{0}^{2} x d x = \frac{x^{2}}{2} ∣_{0}^{2} = 2$ $\Rightarrow \frac{a_{0}}{2} = 1 .$ $a_{n} = \frac{1}{L} \int_{c}^{c + 2 L} f (x) c o s \frac{n π x}{L} d x n = 1, 2, 3, . . .$ $L e t c = 0 . ∴ a_{n} = \frac{1}{1} \int_{0}^{2} x c o s n π x d x$ $a_{n} = \frac{x}{n π} s i n n π x ∣_{0}^{2} - \int_{0}^{2} \frac{1}{n π} s i n n π x d x$ $a_{n} = \frac{1}{n^{2} π^{2}} c o s n π x ∣_{0}^{2} = \frac{1}{n^{2} π^{2}} (c o s 2 n π - 1) = 0$ $b_{n} = \frac{1}{L} \int_{c}^{c + 2 L} f (x) s i n \frac{n π x}{L} d x (n = 1, 2, 3, . . .)$ $T a k e c = 0 . ∴ b_{n} = \frac{1}{1} \int_{0}^{2} x s i n n π x d x$ $b_{n} = \frac{- x c o s n π x}{n π} ∣_{0}^{2} - \int_{0}^{2} \frac{- c o s n π x}{n π} d x$ $b_{n} = \frac{- 2}{n π} + {[\frac{1}{n^{2} π^{2}} s i n n π x]}_{0}^{2}$ $b_{n} = \frac{- 2}{n π} + 0 = \frac{- 2}{n π} (n \neq 0) .$ $∴ i n (1)$ $x = 1 + \underset{n = 1}{\sum^{\infty}} \frac{- 2 s i n n π x}{n π}$ $x = 1 - \frac{2}{π} \underset{n = 1}{\sum^{\infty}} \frac{s i n n π x}{n}$ $L e t x = \frac{3}{π} \notin Z (I f x \in Z, x i s a p o i n t o f$ $d i s c o n t i n u i t y w h o s e o u t p u t i s g i v e n b y$ $\frac{f (x + 0) + f (x - 0)}{2})$ $∴ \frac{3}{π} = 1 - \frac{2}{π} \underset{n = 1}{\sum^{\infty}} \frac{s i n 3 n}{n}$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{s i n 3 n}{n} = \frac{π}{2} (1 - \frac{3}{π})$ $\underset{n = 1}{\sum^{\infty}} \frac{s i n 3 n}{n} = \frac{π - 3}{2}$
	Commented by Tawakalitu. last updated on 15/Aug/16

	$A m v e r y h a p p y . T h a n k y o u s i r . i r e a l l y a p p r e c i a t e .$
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