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JamesJ Group Title

Here's a very nice question. Using the standard power series expansion of e^1, evaluate the following limit ...

  • one year ago
  • one year ago

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  1. JamesJ Group Title
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    \[ \lim_{n \rightarrow \infty} n.\sin(2\pi.e.n!) \]

    • one year ago
  2. amoodarya Group Title
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    |dw:1360717507664:dw|

    • one year ago
  3. JamesJ Group Title
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    Surprisingly, the limit does exist.

    • one year ago
  4. kropot72 Group Title
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    The maximum value of the sine term is unity. Therefore is the limit infinity?

    • one year ago
  5. JamesJ Group Title
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    No

    • one year ago
  6. myko Group Title
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    |dw:1360795091206:dw|

    • one year ago
  7. JamesJ Group Title
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    What you've written is not equal to the limit in question.

    • one year ago
  8. myko Group Title
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    just first step towords it... thinking

    • one year ago
  9. KingGeorge Group Title
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    \[\begin{aligned} \sin(2\pi e n!)&=\sin\left(2\pi \left[\sum_{i=0}^\infty \frac{n!}{i!}\right]\right) \\ &=\sin\left(2\pi\left[\sum_{i=0}^ni! +\sum_{j=n+1}^\infty \frac{n!}{j!}\right]\right) \\ &=\sin\left(\sum_{i=0}^n2\pi i!+\sum_{j=n+1}^\infty \frac{n!}{j!}\right) \\ &=\sin\left(\sum_{i=0}^n 2\pi i!\right)\cos\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right) \\ &\qquad+\sin\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right)\cos\left(\sum_{i=0}^n 2\pi i!\right) \end{aligned}\]Note that the sum is an integer multiple of \(2\pi\), so the cosine of the sum is 1, and the sine term is 0. So we get \[\sin\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right)+\cos\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right)\]Is this on the right track?

    • one year ago
  10. JamesJ Group Title
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    Yes, but you only have one term after your expansion step: the sin(2.pi.sum from n+1 to infinity ...)

    • one year ago
  11. JamesJ Group Title
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    ...and you dropped a 2pi mid way through but picked it up again; that's a small detail.

    • one year ago
  12. KingGeorge Group Title
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    I made a mistake, and had to revise starting from halfway through. I guess that 2pi still slipped through.

    • one year ago
  13. KingGeorge Group Title
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    Anyways, we have\[\sin\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right)\]Then look at\[\lim_{n\to\infty}n\sin\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right)\]Since the sum starts at n+1, the limit of the sum as n goes to infinity is 0. But then we're multiplying by n in the front, so this will take another trick.

    • one year ago
  14. JamesJ Group Title
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    Yes

    • one year ago
  15. kropot72 Group Title
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    The value of the sine term must lie between 0 and plus/minus unity. We are told that the limit is not infinity. Any value of the sine term other than zero would make the limit plus or minus infinity. Therefore the required limit is zero.

    • one year ago
  16. KingGeorge Group Title
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    The limit of the sin term as n goes to infinity is 0, but we are also multiplying by n, so that might change things.

    • one year ago
  17. JamesJ Group Title
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    I'll tell you right now the limit is not zero.

    • one year ago
  18. KingGeorge Group Title
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    If I had to take a guess, I would say either \(e\) or \(\frac{1}{e}\) or some small multiple of one of them.

    • one year ago
  19. JamesJ Group Title
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    George, you're half way there. Hint for the second part as you have it now: use the 'sandwich theorem' (aka 'squeeze theorem'). I prefer sandwiches myself.

    • one year ago
  20. KingGeorge Group Title
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    Alright. I'll give it some thought. Unfortunately, I've got to now. Hopefully I'll be back later with the full solution.

    • one year ago
  21. JamesJ Group Title
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    Great.

    • one year ago
  22. KingGeorge Group Title
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    I think I've finally got it. Notice that for \(n>1\), \[\frac{1}{n+1}\le \sum_{j=n+1}^\infty \frac{n!}{j!}=\frac{1}{(n+1)(n+2)...}\le \sum_{i=0}^\infty \frac{1}{n^i}=\frac{1}{n-1}\]So for sufficiently large \(n\), we have \[n\sin\left(\frac{2\pi}{n+1}\right)\le n\sin\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right)\le n\sin\left(\frac{2\pi}{n-1}\right).\]

    • one year ago
  23. KingGeorge Group Title
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    It's clear that the limit of the least expression is the same as the limit of the greatest expression in that as n goes to infinity. So let's look at the limit of the least.\[\large \lim_{n\to\infty}n\sin\left(\frac{2\pi}{n+1}\right)=\lim_{n\to\infty}\frac{\sin\left(\frac{2\pi}{1/s+1}\right)}{s}\]if \(s=1/n\). Using L'hopitals, we get\[\large\lim_{s\to0}\frac{2\pi\cos\left(\frac{2s\pi}{s+1}\right)}{(s+1)^2}=\frac{2\pi}{1}=2\pi.\]And with that, we've shown that \[\lim_{n\to\infty}n\sin(2\pi en!)=2\pi.\]

    • one year ago
  24. JamesJ Group Title
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    Yes. The way you wrote the inequality above is slightly confusing, but yep! That's the result. It's very surprising, isn't it?

    • one year ago
  25. KingGeorge Group Title
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    Very surprising. Upon first glance it looks very divergent.

    • one year ago
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