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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’.
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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.
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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…
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@Zhao Yunchao – Yeah, there is likely to be more people interested in math problem around you.
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@Brian Lie – Seems like you keep eyes on CMC. Lots of your problems are obtained there.
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Note that π n ! e = π n ! [ 2 + 2 ! 1 + ⋅ ⋅ ⋅ + n ! 1 + ( n + 1 ) ! 1 + ο ( ( n + 1 ) ! 1 ) ] = π a n + n + 1 π + ο ( n + 1 1 ) , where a n is integer. Hence n → ∞ lim π 2 n 2 sin 2 ( π n ! e ) = n → ∞ lim π 2 n 2 sin 2 [ n + 1 π + ο ( n + 1 1 ) ] = n → ∞ lim ( n + 1 ) 2 n 2 ⎣ ⎡ n + 1 π sin ( n + 1 π + ο ( n + 1 1 ) ) ⎦ ⎤ 2 = 1 .