1-find-U-n-0-pi-4-tan-n-tdt-with-n-integr-2-find-lim-n-U-n-3-calculate-n-0-U-n-

Question Number 53471 by maxmathsup by imad last updated on 22/Jan/19

1) f i n d U_{n} = \int_{0}^{\frac{π}{4}} {t a n}^{n} t d t w i t h n i n t e g r .

2) f i n d {l i m}_{n \to + \infty} U_{n}

3) c a l c u l a t e \sum_{n = 0}^{\infty} U_{n}

Commented by Abdo msup. last updated on 23/Jan/19

1) w e h a v e U_{n} = \int_{0}^{\frac{π}{4}} {t a n}^{n - 2} ({t a n}^{2} t + 1 - 1) d t

= \int_{0}^{\frac{π}{4}} (1 + {t a n}^{2} t) {t a n}^{n - 2} t d t - U_{n - 2} a n d b y p a r t s

\int_{0}^{\frac{π}{4}} (1 + {t a n}^{2} t) {t a n}^{n - 2} t d t = {[t a n t {t a n}^{n - 2} t]}_{0}^{\frac{π}{4}}

- \int_{0}^{\frac{π}{4}} t a n t (n - 2) {t a n}^{n - 3} (1 + {t a n}^{2} t) d t

= 1 - (n - 2) \int_{0}^{\frac{π}{4}} {t a n}^{n - 2} (1 + {t a n}^{2} t) d t

= 1 - (n - 2) U_{n - 2} - (n - 2) U_{n} \Rightarrow

U_{n} = 1 - (n - 2) U_{n - 2} - (n - 2) U_{n} - U_{n - 2} \Rightarrow

(n - 1) U_{n} = 1 - (n - 1) U_{n - 2} \Rightarrow

U_{n} = \frac{1}{n - 1} - U_{n - 2} w i t h n ⩾ 2 \Rightarrow

U_{2 n} = \frac{1}{2 n - 1} - U_{2 n - 2} a n d U_{2 n + 1} = \frac{1}{2 n} - U_{2 n - 1}

l e t f i n d U_{2 n} w e h a v e

U_{2 n} + U_{2 n - 2} = \frac{1}{2 n - 1} \Rightarrow

\sum_{k = 1}^{n} {(- 1)}^{k} (U_{2 k} + U_{2 k - 2}) = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k - 1} \Rightarrow

- (U_{2} + U_{0}) + U_{4} + U_{2} - (U_{6} + U_{4}) + \dots

+ {(- 1)}^{n - 1} (U_{2 n - 2} + U_{2 n - 4}) + {(- 1)}^{n} (U_{2 n} + U_{2 n - 2})

= \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k - 1} \Rightarrow {(- 1)}^{n} U_{2 n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k - 1} \Rightarrow

U_{2 n} = {(- 1)}^{n} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k - 1}

Commented by Abdo msup. last updated on 23/Jan/19

l e t f i n d U_{2 n + 1} w e h a v e U_{2 n + 1} = \frac{1}{2 n} - U_{2 n - 1} \Rightarrow

U_{2 n + 1} + U_{2 n - 1} = \frac{1}{2 n} \Rightarrow

\sum_{k = 1}^{n} {(- 1)}^{k} (U_{2 k + 1} + U_{2 k - 1}) = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k} \Rightarrow

- (U_{3} + U_{1}) + U_{5} + U_{3} + \dots . + {(- 1)}^{n} (U_{2 n + 1} + U_{2 n - 1})

= \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k} \Rightarrow

U_{2 n + 1} = \frac{{(- 1)}^{n}}{2} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k} .

Commented by Abdo msup. last updated on 23/Jan/19

w e s e e t h a t (U_{n}) i s d i v e r g e n t b u t w e h a v e

{(- 1)}^{n} U_{2 n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k - 1} =_{k - 1 = p} \sum_{p = 0}^{n - 1} \frac{{(- 1)}^{p + 1}}{2 p + 1}

= - \sum_{p = 0}^{n - 1} \frac{{(- 1)}^{p}}{2 p + 1} \to - \frac{π}{4} (n \to + \infty)

a l s o {(- 1)}^{n} U_{2 n + 1} = \frac{1}{2} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k} \to - \frac{l n (2)}{2} (n \to + \infty)

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jan/19

I_{n} = \int {t a n}^{n - 2} t \times {t a n}^{2} t d t

= \int {t a n}^{n - 2} t ({s e c}^{2} t - 1) d t

= \int {t a n}^{n - 2} {t s e c}^{2} t d t - \int {t a n}^{n - 2} t d t

= \int {t a n}^{n - 2} t \times d (t a n t) - I_{n - 2}

= \frac{{(t a n t)}^{n - 2 + 1}}{(n - 2 + 1)} - I_{n - 2}

s o U_{n} =∣ \frac{{(t a n t)}^{n - 1}}{n - 1} ∣_{0}^{\frac{π}{4}} - U_{n - 2}

U_{n} = \frac{1}{n - 1} - U_{n - 2}