let U_n = ∫_0 ^∞ (dt/((1+t^3 )^n )) dt (n≥1) 1) calculate (U_(n+1) /U_n ) 2) study the serie Σln((U_(n+1) /U_n )) and prove that lim_(n→+∞) U


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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
$1) we have U_n =∫_0 ^∞ ((1+t^3 )/((1+t^3 )^(n+1) )) dt =∫_0 ^∞ (dt/((1+t^3 )^(n+1) )) +∫_0 ^∞ (t^3 /((1+t^3 )^(n+1) )) dt ∫_0 ^∞ (dt/((1+t^3 )^(n+1) )) =U_(n+1) ∫_0 ^∞ (t^3 /((1+t^3 )^(n+1) )) dt = (1/3)∫_0 ^∞ t (3t^2 )(1+t^3 )^(−n−1) dt by parts u =t and v^, =(3t^2 )(1+t^3 )^(−n−1) ⇒ ∫_0 ^∞ (t^3 /((1+t^3 )^(n+1) )) dt = (1/3){ [−(t/n)(1+t^3 )^(−n) ]_0 ^∞ +∫_0 ^∞ (1/n)(1+t^3 )^(−n) dt} =(1/(3n)) ∫_0 ^∞ (dt/((1+t^3 )^n )) =(1/(3n)) U_n ⇒ U_n = U_(n+1) +(1/(3n)) U_n ⇒(1−(1/(3n)))U_n =U_(n+1) ⇒ (((3n−1)/(3n)))U_n =U_(n+1) ⇒(U_(n+1) /U_n ) = 1−(1/(3n))$
$1) w e h a v e U_{n} = \int_{0}^{\infty} \frac{1 + t^{3}}{{(1 + t^{3})}^{n + 1}} d t$ $= \int_{0}^{\infty} \frac{d t}{{(1 + t^{3})}^{n + 1}} + \int_{0}^{\infty} \frac{t^{3}}{{(1 + t^{3})}^{n + 1}} d t$ $\int_{0}^{\infty} \frac{d t}{{(1 + t^{3})}^{n + 1}} = U_{n + 1}$ $\int_{0}^{\infty} \frac{t^{3}}{{(1 + t^{3})}^{n + 1}} d t = \frac{1}{3} \int_{0}^{\infty} t (3 t^{2}) {(1 + t^{3})}^{- n - 1} d t b y p a r t s$ $u = t a n d v^{,} = (3 t^{2}) {(1 + t^{3})}^{- n - 1} \Rightarrow$ $\int_{0}^{\infty} \frac{t^{3}}{{(1 + t^{3})}^{n + 1}} d t =$ $\frac{1}{3} {{[- \frac{t}{n} {(1 + t^{3})}^{- n}]}_{0}^{\infty} + \int_{0}^{\infty} \frac{1}{n} {(1 + t^{3})}^{- n} d t}$ $= \frac{1}{3 n} \int_{0}^{\infty} \frac{d t}{{(1 + t^{3})}^{n}} = \frac{1}{3 n} U_{n} \Rightarrow$ $U_{n} = U_{n + 1} + \frac{1}{3 n} U_{n} \Rightarrow (1 - \frac{1}{3 n}) U_{n} = U_{n + 1} \Rightarrow$ $(\frac{3 n - 1}{3 n}) U_{n} = U_{n + 1} \Rightarrow \frac{U_{n + 1}}{U_{n}} = 1 - \frac{1}{3 n}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum