find Σ_(n=1) ^∞ ((sin(nx))/n).


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Question Number 28614 by abdo imad last updated on 27/Jan/18

$f i n d \sum_{n = 1}^{\infty} \frac{s i n (n x)}{n} .$
Commented by abdo imad last updated on 30/Jan/18

$l e t d e v e l o p p t h e f u n c t i o n f (x) = \frac{x}{2} p e r i o d i c b y 2 π a n d o d d$ $f (x) = \sum_{n = 1}^{\infty} a_{n} s i n (n x) a n d a_{n} = \frac{2}{2 π} \int_{[T]} \frac{x}{2} s i n (n x) d x$ $= \frac{1}{π} \int_{0}^{π} x s i n (n x) d x a n d b y p a r t s$ ${π a_{n} = \frac{- x}{n} c o s (n x)]}_{0}^{π} - \int_{0}^{π} \frac{- 1}{n} c o s (n x) d x$ $= \frac{- π}{n} {(- 1)}^{n} + \frac{1}{n^{2}} {[s i n (n x)]}_{0}^{π} = \frac{π {(- 1)}^{n - 1}}{n} s o$ $a_{n} = \frac{{(- 1)}^{n - 1}}{n} a n d$ $\frac{x}{2} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} s i n (n x) l e t d o t h e c h . = π - t s o$ $\frac{π - t}{2} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} s i n (n π - n t) b u t$ $s i n (n π - n t) = s i n (n π) c o s (n t) - c o s (n π) s i n (n t)$ $= {(- 1)}^{n - 1} s i n (n t) \Rightarrow \frac{π - t}{2} = \sum_{n = 1}^{\infty} \frac{s i n (n t)}{n} f i n a l l y w e h a v e$ $\sum_{n = 1}^{\infty} \frac{s i n (n x)}{n} = \frac{π - x}{2} .$
Commented by abdo imad last updated on 30/Jan/18

$t h e c o n v e r g e n c e o f t h i s s e r i e i s a s s u r e d b y A b e l D i r i c h l e t$ $t h e o r e m l e t r e m e m b e r t h i s t h e o r e m . i f Σ a_{n} (x) v_{n} (x) i s a$ $s e r i e o f f u n c t i o n (a_{n} > 0 a n d v_{n} > 0) / a_{n} d e c r e a s e a n d a_{n} {(x)}_{n \to \infty} \to 0$ $a n d \exists m > 0 / ∣ \sum_{k = n_{0}}^{n} v_{k} (x) ∣⩽ m s o t h e s e r i e c o n v e r g e s .$
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