Evaluate-sin-3n-n-from-1-to-infinity-

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

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, \dots

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, \dots)

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 .