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Question Number 61660 by maxmathsup by imad last updated on 05/Jun/19

l e t U_{n} = \int_{0}^{\infty} \frac{d t}{{(1 + t^{3})}^{n}} d t (n ⩾ 1)

1) c a l c u l a t e \frac{U_{n + 1}}{U_{n}}

2) s t u d y t h e s e r i e Σ l n (\frac{U_{n + 1}}{U_{n}}) a n d p r o v e t h a t {l i m}_{n \to + \infty} U_{n} = 0

Commented by prof Abdo imad last updated on 07/Jun/19

1) w e h a v e U_{n} = \int_{0}^{\infty} \frac{1 + t^{3}}{{(1 + t^{3})}^{n + 1}} d t

= \int_{0}^{\infty} \frac{d t}{{(1 + t^{3})}^{n + 1}} + \int_{0}^{\infty} \frac{t^{3}}{{(1 + t^{3})}^{n + 1}} d t

\int_{0}^{\infty} \frac{d t}{{(1 + t^{3})}^{n + 1}} = U_{n + 1}

\int_{0}^{\infty} \frac{t^{3}}{{(1 + t^{3})}^{n + 1}} d t = \frac{1}{3} \int_{0}^{\infty} t (3 t^{2}) {(1 + t^{3})}^{- n - 1} d t b y p a r t s

u = t a n d v^{,} = (3 t^{2}) {(1 + t^{3})}^{- n - 1} \Rightarrow

\int_{0}^{\infty} \frac{t^{3}}{{(1 + t^{3})}^{n + 1}} d t =

\frac{1}{3} {{[- \frac{t}{n} {(1 + t^{3})}^{- n}]}_{0}^{\infty} + \int_{0}^{\infty} \frac{1}{n} {(1 + t^{3})}^{- n} d t}

= \frac{1}{3 n} \int_{0}^{\infty} \frac{d t}{{(1 + t^{3})}^{n}} = \frac{1}{3 n} U_{n} \Rightarrow

U_{n} = U_{n + 1} + \frac{1}{3 n} U_{n} \Rightarrow (1 - \frac{1}{3 n}) U_{n} = U_{n + 1} \Rightarrow

(\frac{3 n - 1}{3 n}) U_{n} = U_{n + 1} \Rightarrow \frac{U_{n + 1}}{U_{n}} = 1 - \frac{1}{3 n}