let A_n =Σ_(k=0) ^(n−1) sin(((kπ)/(2n))) and B_n =Σ_(k=0) ^(n−1) cos(((kπ)/(2n))) 1) find A_n and B_n interms of n 2)calculate lim_(n→+∞) (A_n /B_n ) 3)calculate ((lim_(n→+∞) A_n )/(lim_(n→+∞ ) B


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Question Number 47857 by maxmathsup by imad last updated on 15/Nov/18

$l e t A_{n} = \sum_{k = 0}^{n - 1} s i n (\frac{k π}{2 n}) a n d B_{n} = \sum_{k = 0}^{n - 1} c o s (\frac{k π}{2 n})$ $1) f i n d A_{n} a n d B_{n} i n t e r m s o f n$ $2) c a l c u l a t e {l i m}_{n \to + \infty} \frac{A_{n}}{B_{n}}$ $3) c a l c u l a t e \frac{{l i m}_{n \to + \infty} A_{n}}{{l i m}_{n \to + \infty} B_{n}} .$
Commented by maxmathsup by imad last updated on 17/Nov/18
$we have B_n +iA_n =Σ_(k=0) ^(n−1) e^(i((kπ)/(2n))) =Σ_(k=0) ^(n−1) (e^(i(π/(2n))) )^k = ((1−e^((iπ)/2) )/(1−e^((iπ)/(2n)) )) =((1−i)/(1−cos((π/(2n)))−isin((π/(2n))))) =((1−i)/(2sin^2 ((π/(4n)))−2isin((π/(4n)))cos((π/(4n))))) =((1−i)/(−2i sin((π/(4n)))(cos((π/(4n)))+i sin((π/(4n)))))) =((i+1)/(2sin((π/(4n))))) e^(−((iπ)/(4n))) =((1+i)/(2sin((π/(4n))))){cos((π/(4n)))−i sin((π/(4n)))} =((cos((π/(4n)))−i sin((π/(4n))) +i cos((π/(4n))) +sin((π/(4n))))/(2sin((π/(4n))))) ⇒ B_n =((cos((π/(4n)))+sin((π/(4n))))/(2sin((π/(4n))))) =(1/(2tan((π/(4n))))) +(1/2) A_n =((cos((π/(4n)))−sin((π/(4n))))/(2sin((π/(4n))))) = (1/(2 tan((π/(4n))))) −(1/2) 2) we have (A_n /B_n ) =((cos((π/(4n)))−sin((π/(4n))))/(cos((π/(4n)))+sin((π/(4n))))) ∼((1−(π^2 /(32n^2 ))−(π/(4n)))/(1−(π^2 /(32n^2 ))+(π/(4n)))) →1 (n→+∞) ⇒ lim_(n→+∞) (A_n /B_n ) =1 .$
$w e h a v e B_{n} + {i A}_{n} = \sum_{k = 0}^{n - 1} e^{i \frac{k π}{2 n}} = \sum_{k = 0}^{n - 1} {(e^{i \frac{π}{2 n}})}^{k} = \frac{1 - e^{\frac{i π}{2}}}{1 - e^{\frac{i π}{2 n}}}$ $= \frac{1 - i}{1 - c o s (\frac{π}{2 n}) - i s i n (\frac{π}{2 n})} = \frac{1 - i}{2 {s i n}^{2} (\frac{π}{4 n}) - 2 i s i n (\frac{π}{4 n}) c o s (\frac{π}{4 n})}$ $= \frac{1 - i}{- 2 i s i n (\frac{π}{4 n}) (c o s (\frac{π}{4 n}) + i s i n (\frac{π}{4 n}))}$ $= \frac{i + 1}{2 s i n (\frac{π}{4 n})} e^{- \frac{i π}{4 n}} = \frac{1 + i}{2 s i n (\frac{π}{4 n})} {c o s (\frac{π}{4 n}) - i s i n (\frac{π}{4 n})}$ $= \frac{c o s (\frac{π}{4 n}) - i s i n (\frac{π}{4 n}) + i c o s (\frac{π}{4 n}) + s i n (\frac{π}{4 n})}{2 s i n (\frac{π}{4 n})} \Rightarrow$ $B_{n} = \frac{c o s (\frac{π}{4 n}) + s i n (\frac{π}{4 n})}{2 s i n (\frac{π}{4 n})} = \frac{1}{2 t a n (\frac{π}{4 n})} + \frac{1}{2}$ $A_{n} = \frac{c o s (\frac{π}{4 n}) - s i n (\frac{π}{4 n})}{2 s i n (\frac{π}{4 n})} = \frac{1}{2 t a n (\frac{π}{4 n})} - \frac{1}{2}$ $2) w e h a v e \frac{A_{n}}{B_{n}} = \frac{c o s (\frac{π}{4 n}) - s i n (\frac{π}{4 n})}{c o s (\frac{π}{4 n}) + s i n (\frac{π}{4 n})} \sim \frac{1 - \frac{π^{2}}{32 n^{2}} - \frac{π}{4 n}}{1 - \frac{π^{2}}{32 n^{2}} + \frac{π}{4 n}} \to 1 (n \to + \infty) \Rightarrow$ ${l i m}_{n \to + \infty} \frac{A_{n}}{B_{n}} = 1 .$
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