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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} = \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