Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
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 .

share|improve this question

marked as duplicate by Davide Giraudo, Ross Millikan, Did, Rahul, J.D. Jul 27 '12 at 21:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1  
Is it $\sin(2 \pi n! e)$? I would think so, but parentheses would be appreciated. –  Ross Millikan Jul 27 '12 at 21:30
    
@Belgi: Huh? $ $ –  Did Jul 27 '12 at 21:30
    
@Begi : I don't think that way it can be done or may be too hard . –  Theorem Jul 27 '12 at 21:32
    
My mistake, my comment was about " but somehow i don't remember", I thought the PO doesn't remember the approximation... –  Belgi Jul 27 '12 at 21:33
    
Nice problem, but I'd also like to see a different approach from the classical one. –  Chris's sis Aug 1 '12 at 13:21

3 Answers 3

Look at the usual series for $e$. Then $n!e$ is an integer plus $$\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots.\tag{$1$}$$ The sum $(1)$ is bigger than $\frac{1}{n+1}$. It is less than $$\frac{1}{n+1}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots,$$ a geometric series with sum $\frac{1}{n}$.

share|improve this answer

Remember that $\mathrm e=\sum\limits_{k=0}^n1/k!+R_n$ with $R_n=\sum\limits_{k=n+1}^{+\infty}1/k!$. Hence $n!\,\mathrm e$ equals an integer plus $n!R_n$. Now $1/(n+1)!\lt R_n\lt1/(n\,n!)$, hence $n!R_n=1/n+o(1/n)$, which is all the precision you need.

share|improve this answer

$$ \begin{align} n!e &=n!\sum_{k=0}^\infty\frac1{k!}\\ &=\sum_{k=0}^n\frac{n!}{k!}+\sum_{k=n+1}^\infty\frac1{k!/n!}\\ &\equiv\frac1{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\dots\pmod{\mathbb{Z}} \end{align} $$ where the last sum is greater than $\frac1{n+1}$ yet less than $\frac1n=\frac1{n+1}+\frac1{(n+1)^2}+\frac1{(n+1)^3}+\dots$

Thus, we have the bounds $$ \frac1{n+1}<n!e-\lfloor n!e\rfloor<\frac1n $$ Therefore, $$ n\sin\left(\frac{2\pi}{n+1}\right)<n\sin(2\pi n!e)<n\sin\left(\frac{2\pi}{n}\right) $$ and by the Squeeze Theorem, we get $$ \lim_{n\to\infty}n\sin(2\pi n!e)=2\pi $$

share|improve this answer
    
One doubt that i have is that why are we not taking account of the integer that we get ? as $n\to \infty$ that can contribute a lot as well right ? so what am i still missing –  Theorem Jul 27 '12 at 21:52
    
@Theorem: $\sin(2\pi n+2\pi x)=\sin(2\pi x)$ for any integer $n$. –  robjohn Jul 27 '12 at 21:54
    
Very silly of me. Thanks. –  Theorem Jul 27 '12 at 21:55
    
@robjohn: very detailed! (+1) –  Chris's sis Aug 1 '12 at 13:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.