let-A-n-k-0-n-1-sin-kpi-2n-and-B-n-k-0-n-1-cos-kpi-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-

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

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 .