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I need to calculate $\lim \sin(2\pi n!e)$.

I put it into Wolfram and saw that it is likely to converge to 0.

Of course this would mean that the fractional part of $n!e$ should be always very close to either 0 or 0.5.

It means that the fluctuation of the fractional part is gonna decrease. But how do I show that it will stop around 0 or 0.5?

Cheers.

EDIT: Solved.

$n!e = k + \frac{n!}{(n+1)!}\left( \frac{1}{1} + \frac{1}{n+2} + \frac{1}{(n+2)(n+3)} + \ldots \right) < k + \frac{n!}{(n+1)!}\left ( 1 + \frac{1}{n+1} + \frac{1}{(n+1)^2} + \frac{1}{(n+1)^3} + \ldots \right) = k + \frac{1}{n} $

where $k \in Z$

Therefore, the fractional part of $n!e$ approaches $0$.

Kuba
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  • What is n approaching in the limit? – Zach466920 Jun 10 '15 at 15:29
  • The limit seems to be 0. The sequence is convergent, but I also need to prove it as well as calculate the limit. – Kuba Jun 10 '15 at 15:30
  • Once again, what is the variable that you have approach a value. I assume it to be n. What value is n approaching? Is n approaching infinity? – Zach466920 Jun 10 '15 at 15:31
  • Use Stirling's approximation for $n!$. – barak manos Jun 10 '15 at 15:31
  • How do you "see that the difference between (n+1)!e and n!e is: k+1/(n+1) for some integer k". The difference is (n+1)e, which is decidedly not rational. – user3697176 Jun 10 '15 at 15:48
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    Use Robert Israel's very useful hint. The "tail" is $\lt \frac{n!}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots\right)$. This is a geometric series. – André Nicolas Jun 10 '15 at 15:50
  • @Zach466920 what other limit could there be to consider? Limit for $2 \to -1$? :P – Zardo Jun 10 '15 at 16:13
  • @Zardo well I know the limit is any number between -1 and 1 and that sin oscillates so as n approaches infinity, the limit tends to anything between -1 and 1 ;) – Zach466920 Jun 10 '15 at 16:17
  • Related: http://math.stackexchange.com/questions/76097/what-is-the-limit-of-n-sin-2-pi-cdot-e-cdot-n-as-n-goes-to-infinity – lab bhattacharjee Jun 10 '15 at 16:27
  • @AndréNicolas, why is there (n+1)! in the denominator there? I recon it should be simply n+1. Then how does this help me? Read my edit :) Thanks! – Kuba Jun 10 '15 at 17:58
  • You left out my $(n+1)!$ in the denominator. Let us look at a particular example, $5!\left(\frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!}+\frac{1}{9!}+\cdots\right)$. We have $\frac{1}{7!}\lt \frac{1}{6! \cdot 6}$. We have $\frac{1}{8!}\lt \frac{1}{6! \cdot 6^2}$. And so on. So the full sum is $\lt \frac{5!}{6!}(1+\frac{1}{6}+\frac{1}{6^2}+\cdots)$. Sum the series and simplify. We get $\frac{1}{5}$. In general we get that the tail is $\lt \frac{1}{n}$, which goes to $0$. – André Nicolas Jun 10 '15 at 18:12
  • Yes, yes, sure Andre. Thanks for your thorough explanation, but I simply made a silly mistake during simplifying the expression. I've done the same calculation properly now and proved that the tail approches 0. Thank you for your time and keep being a good person. – Kuba Jun 10 '15 at 18:14
  • Another way: By the Lagrange form of the remainder, the tail $\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\cdots$ is $\lt \frac{e^1}{(n+1)!}$. – André Nicolas Jun 10 '15 at 18:16
  • Instead of editing your question, move the solution to an answer. – lhf Jun 10 '15 at 18:18
  • So I should post the answer myself and tick it? I'm new here. – Kuba Jun 10 '15 at 18:20

2 Answers2

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Hint: $n!\; e = (integer) + \sum_{k=n+1}^\infty n!/k!$

Robert Israel
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  • Ok, so I wrote this down (using Andre Nicolas' comment):

    $n!\cdot \left( \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \ldots \right) < n!\cdot \left( \frac{1}{n+1} + \frac{1}{(n+1)^2} + \ldots \right) = \frac{(n-1)!}{n+1}$

    I think the RHS thing is divergent though, so does this even help me? If not, could you give me another hint?

     – Kuba Jun 10 '15 at 17:54
  • You've mistranscribed André Nicolas's comment: The tail is $n! \left(\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\cdots\right) = n!\left(\frac{1}{n!}\cdot\frac{1}{n+1}+\frac{1}{n!}\cdot\frac{1}{(n+1)(n+2)}+...\right) = \frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\cdots < \frac{1}{n+1}+\frac{1}{(n+1)^2}+\cdots = \frac{1}{n}$. – Brian Tung Jun 10 '15 at 18:06
  • Oh jesus, of course I wrote total nonsense. Thanks. – Kuba Jun 10 '15 at 18:09
  • Yeah, now it makes more sense, the proof is complete. – Kuba Jun 10 '15 at 18:14
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SOLUTION:

$n!e = k + \frac{n!}{(n+1)!}\left( \frac{1}{1} + \frac{1}{n+2} + \frac{1}{(n+2)(n+3)} + \ldots \right) < k + \frac{n!}{(n+1)!}\left ( 1 + \frac{1}{n+1} + \frac{1}{(n+1)^2} + \frac{1}{(n+1)^3} + \ldots \right) = k + \frac{1}{n} $

where $k \in Z$

Therefore, the fractional part of $n!e$ approaches $0$.

Kuba
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