Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

How to find the limit

Limit[n*Sin[2*Pi*Exp[1]*n!], n -> Infinity]    ?

Mathematica 10 outputs

 Interval[{-\[Infinity], \[Infinity]}]

which is not correct since the one equals $2\pi$, being a known limit.

share|improve this question
2  
Hm.. The output makes sense to me. At least it wouldn't make sense that the limit was 2 Pi. –  Jacob Akkerboom Sep 1 at 21:00
    
@ Jacob Akkerboom: Could you base your ungrounded words? –  user64494 Sep 1 at 21:01
1  
@MarkMcClure Yes, you may assume you're on your iPhone :) –  belisarius Sep 1 at 22:31
1  
The domain of n must be given, otherwiese the Problem is not well defined. Simply consider Limit[Sin[2 pi n],n->oo]. –  Dr. Wolfgang Hintze Sep 1 at 22:42
1  
@ user64494: it's just a simplified example showing the importance of the Domain in which die Limit is to be considered. –  Dr. Wolfgang Hintze Sep 1 at 22:51

1 Answer 1

up vote 10 down vote accepted

The limit is definitely computed correctly. Keep in mind that Limit assumes the variable (n, in this case) is continuous. Thus, this is a specific example of the general fact that $f(x)\sin(g(x))$ oscillates back and forth over the whole real line, whenever $f(x)$ and $g(x)$ both increase to $\infty$ with $x$. A plot verifies this is correct.

Plot[n*Sin[2*Pi*Exp[1]*n!], {n, 0, 10},
  PlotPoints -> 400]

enter image description here

Again, n represents a continuous variable (since that's how Plot works) and $n!$ is computed via the continuous analog of the factorial, namely using the Gamma function.


Now, the discrete limit

$$\lim _{n\to \infty }n \sin (2\pi e n!)=2\pi$$

is a separate issue. Proving this is a bit tricky but can be done as follows. First, it's a pretty basic fact of calculus that

$$\frac{\sin (a x)}{x}\to a$$

as $x\to 0$. It's not too much more work to show that

$$\frac{\sin \left(a x+O\left(x^2\right)\right)}{x}\to a,$$

where $O\left(x^2\right)$ represents an expression that is bounded by a constant times $x^2$. So, now, let's examine the argument $2\pi n! e$ of the sine in the original question. This all hinges on the standard expression of $e$ in terms of an infinite series:

$$ e = \sum_{k=0}^{\infty} \frac{1}{k!}. $$

Thus,

\begin{align} 2\pi n!e &= 2\pi n! \sum _ {k=0}^{\infty } \frac{1}{k!}=2\pi \left(\sum _ {k=0}^n \frac{n!}{k!}+\frac{1}{n+1}+\sum _ {k=n+2}^{\infty } \frac{n!}{k!}\right) \\ &= 2\pi\left(M+\frac{1}{n}+O\left(\frac{1}{n^2}\right)\right)=\frac{2\pi }{n}+O\left(\frac{1}{n^2}\right)\mod 2\pi \end{align}

Taking $x=1/n$, we get the desired result.

Of course, this can be tested numerically using beli's plot, though I'd prefer ListPlot to ListLinePlot, since this is truly a discrete phenomenon.

ListPlot[Block[{$MaxExtraPrecision = 1000}, 
  Table[N[n*Sin[2*Pi*E*n!], 100], {n, 0, 400}]],
  Epilog -> {Dashed, Line[{{0, 2 Pi}, {400, 2 Pi}}]}]

enter image description here

share|improve this answer
    
Limit does not handle integer assumptions (as in discrete variables). –  Daniel Lichtblau Sep 2 at 1:15
    
@DanielLichtblau That is what I thought. But, then, how do I explain the fact that Limit[n^2 Sin[2*n*Pi], n->Infinity, Assumptions->Element[n, Integers]]==0? –  Mark McClure Sep 2 at 1:23
    
I suppose that Assumptions in Limit might simply be applied to simplify the expression ahead of the computation. That would explain the difference in the results and be dismissed as designed. A genuine domain restriction would be more properly dealt with as a third argument. –  Mark McClure Sep 2 at 10:40
    
Right on all counts. I should have stated that Limit does not do anything beyond what Simplify or Refine1 might do in terms of handling integrality assumptions. –  Daniel Lichtblau Sep 2 at 14:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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