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What is the limit of $n \sin (2 \pi \cdot e \cdot n!)$ as $n$ goes to infinity?

In order to solve the following limit $$\lim_{n\to\infty} n\sin2\pi n!e$$ . This question is very likely to have been asked.

I remember this question and the answer is like $2\pi$ or something .

I also do remember approximating $n!e$ but somehow i don't remember and can't figure out right now .

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    Is it $\sin(2 \pi n! e)$? I would think so, but parentheses would be appreciated.2012-07-27
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    @Belgi: Huh? $ $2012-07-27
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    @Begi : I don't think that way it can be done or may be too hard .2012-07-27
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    My mistake, my comment was about " but somehow i don't remember", I thought the PO doesn't remember the approximation...2012-07-27
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    Nice problem, but I'd also like to see a different approach from the classical one.2012-08-01

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