study-the-convergence-of-U-n-2-pi-0-pi-2-sinx-1-n-n-

Question Number 50426 by Abdo msup. last updated on 16/Dec/18

s t u d y t h e c o n v e r g e n c e o f

U_{n} = {(\frac{2}{π} \int_{0}^{\frac{π}{2}} {(s i n x)}^{\frac{1}{n}})}^{n}

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

{\frac{1}{π} \times 2 \int_{0}^{\frac{π}{2}} s i n x)}^{2 \times (\frac{1 + n}{2 n}) - 1} {(c o s x)}^{2 \times \frac{1}{2} - 1} d x

f o r m u l a 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} {x c o s}^{2 q - 1} x d x

= \frac{⌈ (p) ⌈ q)}{⌈ p + q)}

s o \frac{1}{π} \times \frac{⌈ (\frac{1 + n}{2 n}) \times ⌈ (\frac{1}{2})}{⌈ (\frac{1 + n}{2 n} + \frac{1}{2})}

= \frac{1}{\sqrt{π}} \times \frac{⌈ (\frac{1}{2 n} + \frac{1}{2})}{⌈ (\frac{1}{2 n} + 1)}

w a i t \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

A n o t h e r a p p r o a c h \dots

1 ⩾ s i n x ⩾ 0

1 ⩾ {(s i n x)}^{\frac{1}{n}} ⩾ 0

\frac{2}{π} \times \int_{0}^{\frac{π}{2}} 1 \times d x ⩾ \frac{2}{π} \int_{0}^{\frac{π}{2}} {(s i n x)}^{\frac{1}{n}} d x ⩾ \frac{2}{π} \int_{0}^{\frac{π}{2}} 0 \times d x

2 ⩾ \frac{2}{π} \int_{0}^{\frac{π}{2}} {(s i n x)}^{\frac{1}{n}} d x ⩾ 0

2^{n} ⩾ {[\frac{2}{π} \int_{0}^{\frac{π}{2}} {(s i n x)}^{\frac{1}{n}} d x]}^{n} ⩾ 0