let f_n (x) =((sin(2(n+1)x))/(sinx)) if x∈]0,(π/2)] and f_n (0)=2(n+1) let u_n = ∫_0 ^(π/2) f_n (x)dx 1) prove that ∀n fromN u_(n+1) −u_n =2(((−1)^(n+1) )/(2n+3)) 2)find lim_(n→+∞) u


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Question Number 41515 by maxmathsup by imad last updated on 08/Aug/18

$l e t f_{n} (x) = \frac{s i n (2 (n + 1) x)}{s i n x} i f x \in] 0, \frac{π}{2}] a n d f_{n} (0) = 2 (n + 1) l e t$ $u_{n} = \int_{0}^{\frac{π}{2}} f_{n} (x) d x$ $1) p r o v e t h a t \forall n f r o m N u_{n + 1} - u_{n} = 2 \frac{{(- 1)}^{n + 1}}{2 n + 3}$ $2) f i n d {l i m}_{n \to + \infty} u_{n}$
Commented by maxmathsup by imad last updated on 12/Aug/18
$1) u_(n+1) −u_n =∫_0 ^(π/2) ((sin(2(n+2)x)−sin(2(n+1)x))/(sinx))dx but sinp−sinq =cos((π/2)−p) +cos((π/2)+q) =2 cos(((π−p+q)/2))cos(((−p−q)/2)) = 2cos((π/2) −((p−q)/2))cos(((p+q)/2))=2sin(((p−q)/2))cos(((p+q)/2)) ⇒ sin(2(n+2)x)−sin(2(n+1)x)=2sin(x)cos((((4n+6)x)/2))⇒ u_(n+1) −u_n =2∫_0 ^(π/2) sin(x)cos((2n+3)x) .(1/(sinx)) dx = 2∫_0 ^(π/2) cos{(2n+3)x}dx =2[(1/(2n +3)) sin{(2n+3)x}]_0 ^(π/2) =(2/(2n+3))sin{(2n+3)(π/2)} =(2/(2n+3)) sin(nπ +2π−(π/2)) =−(2/(2n +3)) sin((π/2) −nπ) =−(2/(2n+3))cos(nπ) =2((−(−1)^n )/(2n+3)) =2(((−1)^(n+1) )/(2n+3)) ⇒ u_(n+1) −u_n =2(((−1)^(n+1) )/(2n+3))$
$1) u_{n + 1} - u_{n} = \int_{0}^{\frac{π}{2}} \frac{s i n (2 (n + 2) x) - s i n (2 (n + 1) x)}{s i n x} d x b u t$ $s i n p - s i n q = c o s (\frac{π}{2} - p) + c o s (\frac{π}{2} + q) = 2 c o s (\frac{π - p + q}{2}) c o s (\frac{- p - q}{2})$ $= 2 c o s (\frac{π}{2} - \frac{p - q}{2}) c o s (\frac{p + q}{2}) = 2 s i n (\frac{p - q}{2}) c o s (\frac{p + q}{2}) \Rightarrow$ $s i n (2 (n + 2) x) - s i n (2 (n + 1) x) = 2 s i n (x) c o s (\frac{(4 n + 6) x}{2}) \Rightarrow$ $u_{n + 1} - u_{n} = 2 \int_{0}^{\frac{π}{2}} s i n (x) c o s ((2 n + 3) x) . \frac{1}{s i n x} d x$ $= 2 \int_{0}^{\frac{π}{2}} c o s {(2 n + 3) x} d x = 2 {[\frac{1}{2 n + 3} s i n {(2 n + 3) x}]}_{0}^{\frac{π}{2}}$ $= \frac{2}{2 n + 3} s i n {(2 n + 3) \frac{π}{2}} = \frac{2}{2 n + 3} s i n (n π + 2 π - \frac{π}{2})$ $= - \frac{2}{2 n + 3} s i n (\frac{π}{2} - n π) = - \frac{2}{2 n + 3} c o s (n π) = 2 \frac{- {(- 1)}^{n}}{2 n + 3} = 2 \frac{{(- 1)}^{n + 1}}{2 n + 3} \Rightarrow$ $u_{n + 1} - u_{n} = 2 \frac{{(- 1)}^{n + 1}}{2 n + 3}$
Commented by maxmathsup by imad last updated on 12/Aug/18

$2) w e h a v e \sum_{k = 0}^{n - 1} (u_{k + 1} - u_{k}) = 2 \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k + 1}}{2 k + 3} \Rightarrow$ $u_{n} - u_{0} = 2 \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k + 1}}{2 k + 3} (j = k + 1)$ $= 2 \sum_{j = 1}^{n} \frac{{(- 1)}^{j}}{2 j + 1} \Rightarrow {l i m}_{n \to + \infty} u_{n} = u_{0} + 2 \sum_{j = 1}^{\infty} \frac{{(- 1)}^{j}}{2 j + 1}$ $= u_{0} + 2 \sum_{j = 0}^{\infty} \frac{{(- 1)}^{j}}{2 j + 1} - 2 = u_{0} - 2 + 2 \frac{π}{4} = \frac{π}{2} + u_{0} - 2 b u t$ $u_{0} = \int_{0}^{\frac{π}{2}} 2 d x = π \Rightarrow {l i m}_{n \to + \infty} u_{n} = \frac{3 π}{2} - 2 .$
Commented by math khazana by abdo last updated on 13/Aug/18

$u_{0} = \int_{0}^{\frac{π}{2}} \frac{s i n (2 x)}{s i n x} d x = 2 \int_{0}^{\frac{π}{2}} c o s x d x = 2 \Rightarrow$ ${l i m}_{n \to + \infty} u_{n} = 2 - 2 + \frac{2 π}{4} \Rightarrow$ ${l i m}_{n \to + \infty} u_{n} = \frac{π}{2} .$
	Answered by alex041103 last updated on 10/Aug/18

	$W e k n o w t h a t :$ $s i n (2 (n + 2) x) = s i n (2 (n + 1) x + 2 x) =$ $= c o s (2 x) s i n (2 (n + 1) x) + s i n (2 x) c o s (2 (n + 1) x)$ $c o s (2 x) = 1 - 2 {s i n}^{2} x$ $s i n (2 x) = 2 s i n (x) c o s (x)$ $\Rightarrow u_{n + 1} = \int_{0}^{π / 2} \frac{(1 - 2 {s i n}^{2} x) s i n (2 (n + 1) x)}{s i n (x)} d x + \int_{0}^{π / 2} \frac{2 s i n (x) c o s (x) c o s (2 (n + 1) x)}{s i n (x)} d x$ $= u_{n} + 2 \int_{0}^{π / 2} (c o s (x) c o s (2 (n + 1) x) - s i n (x) s i n (2 (n + 1) x)) d x =$ $\Rightarrow u_{n + 1} - u_{n} = 2 \int c o s ((2 n + 3) x) d x =$ $= \frac{2}{2 n + 3} \int c o s ((2 n + 3) x) d ((2 n + 3) x) =$ $= \frac{2}{2 n + 1} [s i n ((2 n + 3) \frac{π}{2}) - s i n (0)]$ $s i n ((2 n + 3) \frac{π}{2}) = s i n ((n + 1) π + \frac{π}{2})$ $s i n (t + π) = c o s (π) s i n (t) + s i n (π) c o s (t) =$ $= - s i n (t)$ $\Rightarrow s i n ((n + 1) π + \frac{π}{2}) = {(- 1)}^{n + 1} s i n (π / 2) = {(- 1)}^{n + 1}$ $\Rightarrow u_{n + 1} - u_{n} = \frac{2 {(- 1)}^{n + 1}}{2 n + 3}$ $\Rightarrow \underset{k = 0}{\sum^{n}} u_{k + 1} - u_{k} = u_{n + 1} - u_{0} = 2 \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k + 1}}{2 k + 3}$ $u_{0} = \int_{0}^{π / 2} \frac{s i n (2 x)}{s i n (x)} d x = 2 \int_{0}^{π / 2} c o s (x) d x = 2$ $\Rightarrow u_{n} = 2 \underset{k = - 1}{\sum^{n - 1}} \frac{{(- 1)}^{k + 1}}{2 k + 3} = 2 \underset{k = 0 - 1}{\sum^{n - 1}} \frac{{(- 1)}^{k + 1}}{2 (k + 1) + 1}$ $\Rightarrow u_{n} = \underset{k = 0}{\sum^{n}} \frac{2 {(- 1)}^{k}}{2 k + 1}$ $Double subscripts: use braces to clarify$ $A n s . u_{\infty} = π / 2$
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