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Question Number 55571 by maxmathsup by imad last updated on 26/Feb/19

l e t u_{n} = \int_{\frac{π}{n + 1}}^{\frac{π}{n}} \sqrt{t a n (x)} d x w i t h n ⩾ 3

1) c a l c u l a t e U_{n} i n t e r m s o f n a n d c a l c u l a t e {l i m}_{n \to + \infty} U_{n}

2) f i n d n a t u r e o f t h e s e r i e \sum_{n ⩾ 3} U_{n}

Commented by maxmathsup by imad last updated on 03/Mar/19

1) w e h a v e p r o v e d t h a t

\int \sqrt{t a n x} d x = \frac{1}{\sqrt{2}} {l n (\sqrt{\frac{t a n x - \sqrt{2 t a n x} + 1}{t a n x + \sqrt{2 t a n x} + 1}} + a r c t a n (\sqrt{2 t a n x} + 1) + a r c t a n (\sqrt{2 t a n x} - 1)} + c

{\Rightarrow u_{n} = \frac{1}{\sqrt{2}} [l n \sqrt{\frac{t a n n x - \sqrt{2 t a n x} + 1}{t a n x + \sqrt{2 t a n x} + 1}}) + a r c t a n (\sqrt{2 t a n x} + 1) + a r c t a n (\sqrt{2 t a n x} - 1)]}_{\frac{π}{n + 1}}^{\frac{π}{n}}

= \frac{1}{\sqrt{2}} A_{n} - \frac{1}{\sqrt{2}} B_{n} w i t h

A_{n} = l n (\sqrt{\frac{t a n (\frac{π}{n}) - \sqrt{2 t a n (\frac{π}{n})} + 1}{t a n (\frac{π}{n}) + \sqrt{2 t a n (\frac{π}{n})} + 1}}) + a r c t a n (\sqrt{2 t a n (\frac{π}{n})} + 1)

+ a r c t a n (\sqrt{2 t a n (\frac{π}{n})} - 1)} a n d

B_{n} = l n (\sqrt{\frac{t a n (\frac{π}{n + 1}) - \sqrt{2 t a n (\frac{π}{n + 1})} + 1}{t a n (\frac{π}{n + 1}) + \sqrt{2 t a n (\frac{π}{n + 1})} + 1}}) + a r c t a n (\sqrt{2 t a n (\frac{π}{n + 1})} + 1)

+ a r c t a n (\sqrt{2 t a n (\frac{π}{n + 1})} - 1}

w e h a v e {l i m}_{n \to + \infty} A_{n} = l n (\frac{1}{1}) + a r c t a n (1) + a r c t a n (- 1) = 0

{l i m}_{n \to + \infty} B_{n} = l n (\frac{1}{1}) + a r c t a n (1) + a r c t a n (- 1) = 0 \Rightarrow {l i m}_{n \to + \infty} u_{n} = 0