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Question Number 55907 by gunawan last updated on 06/Mar/19

for every n \in N, f_{n} (x) = n x {(1 - x^{2})}^{n},

for every x, 0 ⩽ x ⩽ 1

and a_{n} = \int_{0}^{1} f_{n} (x) d x .

If S_{n} = \sin (π a_{n}), for every

n \in N, then \lim_{n \to \infty} s_{n} = \dots

Commented by maxmathsup by imad last updated on 06/Mar/19

w e h a v e a_{n} = \int_{0}^{1} n x {(1 - x^{2})}^{n} d x \Rightarrow π a_{n} = n π \int_{0}^{1} x {(1 - x^{2})}^{n} d x b u t

\int_{0}^{1} x {(1 - x^{2})}^{n} d x = - \frac{1}{2 (n + 1)} {[{(1 - x^{2})}^{n + 1}]}_{0}^{1} = \frac{1}{2 n + 2} \Rightarrow π a_{n} = \frac{n π}{2 n + 2}

= \frac{n π}{2 n (1 + \frac{1}{n})} = \frac{π}{2 (1 + \frac{1}{n})} \sim \frac{π}{2} (1 - \frac{1}{n}) (n \to + \infty \Rightarrow s i n (π a_{n}) \sim s i n (\frac{π}{2} - \frac{π}{2 n})

\Rightarrow {l i m}_{n \to + \infty} S_{n} = s i n (\frac{π}{2}) = 1 .

Answered by 121194 last updated on 06/Mar/19

a_{n} = \underset{0}{\int^{1}} n x {(1 - x^{2})}^{n} d x

y = 1 - x^{2} \Rightarrow d y = - 2 x d x \Rightarrow x d x = - \frac{d y}{2}

x = 0 \Rightarrow y = 1

x = 1 \Rightarrow y = 0

a_{n} = n \underset{0}{\int^{1}} {(1 - x^{2})}^{n} x d x = - \frac{n}{2} \underset{1}{\int^{0}} y^{n} d y = \frac{n}{2} \underset{0}{\int^{1}} y^{n} d y

= \frac{n}{2 (n + 1)}

\underset{n \to \infty}{\lim s i n} [\frac{π n}{2 (n + 1)}] = \sin (\frac{π}{2} \lim_{n \to \infty} \frac{n}{n + 1}) = \sin \frac{π}{2} = 1

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