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JamesJ

  • 2 years ago

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

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  1. JamesJ
    • 2 years ago
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    \[ \lim_{n \rightarrow \infty} n.\sin(2\pi.e.n!) \]

  2. amoodarya
    • 2 years ago
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    |dw:1360717507664:dw|

  3. JamesJ
    • 2 years ago
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    Surprisingly, the limit does exist.

  4. kropot72
    • 2 years ago
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    The maximum value of the sine term is unity. Therefore is the limit infinity?

  5. JamesJ
    • 2 years ago
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    No

  6. myko
    • 2 years ago
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    |dw:1360795091206:dw|

  7. JamesJ
    • 2 years ago
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    What you've written is not equal to the limit in question.

  8. myko
    • 2 years ago
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    just first step towords it... thinking

  9. KingGeorge
    • 2 years ago
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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?

  10. JamesJ
    • 2 years ago
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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 ...)

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

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

  13. KingGeorge
    • 2 years ago
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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.

  14. JamesJ
    • 2 years ago
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    Yes

  15. kropot72
    • 2 years ago
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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.

  16. KingGeorge
    • 2 years ago
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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.

  17. JamesJ
    • 2 years ago
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    I'll tell you right now the limit is not zero.

  18. KingGeorge
    • 2 years ago
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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.

  19. JamesJ
    • 2 years ago
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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.

  20. KingGeorge
    • 2 years ago
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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.

  21. JamesJ
    • 2 years ago
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    Great.

  22. KingGeorge
    • 2 years ago
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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 \[n\sin\left(\frac{2\pi}{n+1}\right)\le n\sin\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right)\le n\sin\left(\frac{2\pi}{n-1}\right).\]

  23. KingGeorge
    • 2 years ago
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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 \[\lim_{n\to\infty}n\sin(2\pi en!)=2\pi.\]

  24. JamesJ
    • 2 years ago
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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?

  25. KingGeorge
    • 2 years ago
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    Very surprising. Upon first glance it looks very divergent.

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