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Question Number 61530 by maxmathsup by imad last updated on 04/Jun/19

l e t U_{n} = \int_{0}^{\infty} \frac{x^{- 2 n}}{1 + x^{4}} d x w i t h n i n t e g r n a t u r a l a n d n ⩾ 1

1) c a l c u l a t e U_{n} i n t e r m s o f n

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

3) s t u d y t h e s e r i e Σ U_{n}

Commented by maxmathsup by imad last updated on 04/Jun/19

1) c h a n g e m e n t x^{4} = t g i v e x = t^{\frac{1}{4}} \Rightarrow U_{n} = \int_{0}^{\infty} \frac{{(t^{\frac{1}{4}})}^{\frac{- 2}{n}}}{1 + t} \frac{1}{4} t^{\frac{1}{4} - 1} d t

= \frac{1}{4} \int_{0}^{\infty} \frac{t^{- \frac{1}{2 n} + \frac{1}{4} - 1}}{1 + t} d t = \frac{1}{4} \int_{0}^{\infty} \frac{t^{\frac{1}{4} - \frac{1}{2 n} - 1}}{1 + t} d t l e t a = \frac{1}{4} - \frac{1}{2 n} \Rightarrow a = \frac{2 n - 4}{8 n} = \frac{n - 2}{4 n} > 0

f o r n > 2 a n d a - 1 = \frac{n - 2}{4 n} - 1 = \frac{n - 2 - 4 n}{4 n} = \frac{- 3 n - 2}{4 n} < 0 \Rightarrow 0 < a < 1 \Rightarrow

U_{n} = \frac{1}{4} \frac{π}{s i n (π a)} = \frac{π}{4 s i n (π (\frac{1}{4} - \frac{1}{2 n}))} = \frac{π}{4 s i n (\frac{π}{4} - \frac{π}{2 n})}

Commented by maxmathsup by imad last updated on 04/Jun/19

f o r g i v e U_{n} = \int_{0}^{\infty} \frac{x^{- \frac{2}{n}}}{1 + x^{4}} d x w i t h n ⩾ 3

Commented by maxmathsup by imad last updated on 04/Jun/19

i h a v e u s e d t h e r e s u l t \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)} w i t h 0 < a < 1

Commented by maxmathsup by imad last updated on 04/Jun/19

2) s i n (\frac{π}{4} - \frac{π}{2 n}) = s i n (\frac{π}{4}) c o s (\frac{π}{2 n}) - c o s (\frac{π}{4}) s i n (\frac{π}{2 n}) = \frac{\sqrt{2}}{2} (c o s (\frac{π}{2 n}) - s i n (\frac{π}{2 n})) \Rightarrow

c o s (\frac{π}{2 n}) \sim 1 - \frac{{(\frac{π}{2 n})}^{2}}{2} = 1 - \frac{π^{2}}{8 n^{2}}, s i n (\frac{π}{2 n}) \sim \frac{π}{2 n} \Rightarrow

U_{n} \sim \frac{π}{2 \sqrt{2} {1 - \frac{π^{2}}{8 n^{2}} - \frac{π}{2 n}}} = \frac{1}{2 \sqrt{2} {\frac{1}{π} - \frac{π}{8 n^{2}} - \frac{1}{2 n}}} \Rightarrow n^{2} U_{n} \sim \frac{n^{2}}{2 \sqrt{2} {\frac{1}{π} - \frac{π}{8 n^{2}} - \frac{1}{2 n}}}

{l i m}_{n \to + \infty} (\frac{1}{π} - \frac{π}{8 n^{2}} \frac{1}{2 n}) = \frac{1}{π} \Rightarrow {l i m}_{n \to + \infty} n^{2} U_{n} = + \infty

Commented by maxmathsup by imad last updated on 04/Jun/19

3) w e {l i m}_{n \to \infty} U_{n} = \frac{π}{2 \sqrt{2}} \neq 0 \Rightarrow Σ U_{n} d i v e r g e s .