let-U-n-1-n-2-n-x-1-x-dx-with-n-3-1-calculate-and-determine-lim-n-U-n-2-study-the-convergence-of-U-n-

Question Number 65004 by mathmax by abdo last updated on 24/Jul/19

l e t U_{n} = \int_{\frac{1}{n}}^{\frac{2}{n}} Γ (x) Γ (1 - x) d x w i t h n ⩾ 3

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

2) s t u d y t h e c o n v e r g e n c e o f Σ U_{n}

Commented by mathmax by abdo last updated on 24/Jul/19

1) w e h a v e Γ (x) . Γ (1 - x) = \frac{π}{s i n (π x)} \Rightarrow U_{n} = \int_{\frac{1}{n}}^{\frac{2}{n}} \frac{π}{s i n (π x)} d x

=_{π x = t} π \int_{\frac{π}{n}}^{\frac{2 π}{n}} \frac{d t}{π s i n t} = \int_{\frac{π}{n}}^{\frac{2 π}{n}} \frac{d t}{s i n t} =_{t a n (\frac{t}{2}) = u} \int_{t a n (\frac{π}{2 n})}^{t a n (\frac{π}{n})} \frac{1}{\frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= \int_{t a n (\frac{π}{2 n})}^{t a n (\frac{π}{n})} \frac{d u}{u} = {[l n ∣ u ∣]}_{t a n (\frac{π}{2 n})}^{t a n (\frac{π}{n})} = l n ∣ \frac{t a n (\frac{π}{n})}{t a n (\frac{π}{2 n})} ∣ \Rightarrow U_{n} = l n ∣ \frac{t a n (\frac{π}{n})}{t a n (\frac{π}{2 n})} ∣

w e h a v e t a n (\frac{π}{n}) \sim \frac{π}{n} a n d t a n (\frac{π}{2 n}) \sim \frac{π}{2 n} \Rightarrow

\frac{t a n (\frac{π}{n})}{t a n (\frac{π}{2 n})} \sim \frac{π}{n} \frac{2 n}{π} = 2 \Rightarrow {l i m}_{n \to + \infty} U_{n} = l n (2)

2) {l i m}_{n \to + \infty} U_{n} \neq 0 \Rightarrow Σ U_{n} d i v e r g e s .