lim-n-sin-2-pi-n-2-n-

Question Number 145531 by mathdanisur last updated on 05/Jul/21

{\underset{n \to \infty}{l i m s i n}}^{2} π \sqrt{n^{2} + n} = ?

Answered by Olaf_Thorendsen last updated on 05/Jul/21

\sin^{2} π \sqrt{n^{2} + n} = \sin^{2} π n \sqrt{1 + \frac{1}{n}}

\sin^{2} π \sqrt{n^{2} + n} \underset{\infty}{\sim} \sin^{2} π n (1 + \frac{1}{2 n})

\sin^{2} π \sqrt{n^{2} + n} \underset{\infty}{\sim} \sin^{2} (π n + \frac{π}{2})

\sin^{2} π \sqrt{n^{2} + n} \underset{\infty}{\sim} \cos^{2} (π n) = {({(- 1)}^{n})}^{2} = 1

Commented by mathdanisur last updated on 05/Jul/21

c o o l t h a n k s S e r

Commented by mathdanisur last updated on 06/Jul/21

S e r, c a n t h e a n s w e r 0 ?

Answered by mathmax by abdo last updated on 05/Jul/21

we have π \sqrt{n^{2} + n} = n π \sqrt{1 + \frac{1}{n}} = n π {(1 + \frac{1}{n})}^{\frac{1}{2}}

\sim n π {1 + \frac{1}{2 n} + \frac{(\frac{1}{2}) (\frac{1}{2} - 1)}{{2 n}^{2}} + o (\frac{1}{n^{3}})}

\sim n π + \frac{π}{2} - \frac{π}{8 n} + o (\frac{1}{n^{2}}) \Rightarrow

\sin (π \sqrt{n^{2} + n}) \sim \sin (n π + \frac{π}{2} - \frac{π}{8 n} + o (\frac{1}{n^{2}})) \sim {(- 1)}^{n} \Rightarrow

\sin^{2} (π \sqrt{n^{2} + n}) \sim 1 \Rightarrow \lim_{n \to + \infty} \sin^{2} π \sqrt{n^{2} + n} = 1

Commented by mathdanisur last updated on 05/Jul/21

c o o l t h a n k s S e r

Commented by mathdanisur last updated on 06/Jul/21

S e r, c a n t h e a n s w e r 0 ?

Answered by Dwaipayan Shikari last updated on 05/Jul/21

{s i n}^{2} (π \sqrt{{(n + \frac{1}{2})}^{2} - \frac{1}{4}})

{s i n}^{2} (π (n + \frac{1}{2}) \sqrt{1 - \frac{1}{{(2 n + 1)}^{2}}})

\lim_{n \to \infty} \sin^{2} (\frac{π}{2} (2 n + 1) (1 - \frac{1}{2 {(2 n + 1)}^{2}}))

= \sin^{2} (π n + \frac{π}{2} - \frac{1}{2 {(2 n + 1)}^{2}}) \approx \sin^{2} (π n + \frac{π}{2})

= {(\pm 1)}^{2} = 1

Commented by mathdanisur last updated on 05/Jul/21

c o o l t h a n k s S e r

Commented by mathdanisur last updated on 06/Jul/21

S e r, c a n t h e a n s w e r 0 ?