Interesting Limit (2)

Calculus Level 2

lim n n 2 sin 2 ( π n ! e ) π 2 = ? \large\lim\limits_{n\rightarrow\infty}\frac{n^2\sin^2(\pi n!e)}{\pi^2}=\, ?


The answer is 1.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Brian Lie
Mar 6, 2018

Note that π n ! e = π n ! [ 2 + 1 2 ! + + 1 n ! + 1 ( n + 1 ) ! + ο ( 1 ( n + 1 ) ! ) ] = π a n + π n + 1 + ο ( 1 n + 1 ) , \begin{aligned} \large\pi n!e&=\large\pi n!\left[2+\frac{1}{2!}+···+\frac{1}{n!}+\frac{1}{(n+1)!}+\omicron\left(\frac{1}{(n+1)!}\right)\right]\\ &=\large\pi a_n+\frac{\pi}{n+1}+\omicron\left(\frac{1}{n+1}\right), \end{aligned} where a n a_n is integer. Hence lim n n 2 sin 2 ( π n ! e ) π 2 = lim n n 2 π 2 sin 2 [ π n + 1 + ο ( 1 n + 1 ) ] = lim n n 2 ( n + 1 ) 2 [ sin ( π n + 1 + ο ( 1 n + 1 ) ) π n + 1 ] 2 = 1. \begin{aligned} \large\lim\limits_{n\rightarrow\infty}\frac{n^2\sin^2(\pi n!e)}{\pi^2}&=\large\lim\limits_{n\rightarrow\infty}\frac{n^2}{\pi^2}\sin^2\left[\frac{\pi}{n+1}+\omicron\left(\frac{1}{n+1}\right)\right]\\ &=\large\lim\limits_{n\rightarrow\infty}\frac{n^2}{(n+1)^2}\left[\frac{\sin\left(\frac{\pi}{n+1}+\omicron\left(\frac{1}{n+1}\right)\right)}{\frac{\pi}{n+1}}\right]^2\\ &=\large1. \end{aligned}

I’m sorry...but I think you have to discuss n. When n is a large odd, ‘an’ is even, and above derivation is correct. If n is even, then ‘an’ is odd and the limit is -1. So the answer should be ‘it doesn’t exist’.

Zhao Yunchao - 3 years, 3 months ago

Log in to reply

Thank you for correcting it and I have edited it again. By the way, you're the first Chinese I've ever seen on Brilliant except me.

Brian Lie - 3 years, 3 months ago

Log in to reply

Aha, Really? I have invited some freshmen to use Brilliant in recent days. Perhaps they answer questions more rather than comment on solutions,which is same to most Chinese visitors…

Zhao Yunchao - 3 years, 3 months ago

Log in to reply

@Zhao Yunchao Yeah, there is likely to be more people interested in math problem around you.

Brian Lie - 3 years, 3 months ago

Log in to reply

@Brian Lie Seems like you keep eyes on CMC. Lots of your problems are obtained there.

Zhao Yunchao - 3 years, 3 months ago

Log in to reply

@Zhao Yunchao Yes, you're right!

Brian Lie - 3 years, 3 months ago

1 pending report

Vote up reports you agree with

×

Problem Loading...

Note Loading...

Set Loading...