JamesJ
  • JamesJ
Here's a very nice question. Using the standard power series expansion of e^1, evaluate the following limit ...
Mathematics
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
JamesJ
  • JamesJ
\[ \lim_{n \rightarrow \infty} n.\sin(2\pi.e.n!) \]
amoodarya
  • amoodarya
|dw:1360717507664:dw|
JamesJ
  • JamesJ
Surprisingly, the limit does exist.

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kropot72
  • kropot72
The maximum value of the sine term is unity. Therefore is the limit infinity?
JamesJ
  • JamesJ
No
anonymous
  • anonymous
|dw:1360795091206:dw|
JamesJ
  • JamesJ
What you've written is not equal to the limit in question.
anonymous
  • anonymous
just first step towords it... thinking
KingGeorge
  • KingGeorge
\[\begin{aligned} \sin(2\pi e n!)&=\sin\left(2\pi \left[\sum_{i=0}^\infty \frac{n!}{i!}\right]\right) \\ &=\sin\left(2\pi\left[\sum_{i=0}^ni! +\sum_{j=n+1}^\infty \frac{n!}{j!}\right]\right) \\ &=\sin\left(\sum_{i=0}^n2\pi i!+\sum_{j=n+1}^\infty \frac{n!}{j!}\right) \\ &=\sin\left(\sum_{i=0}^n 2\pi i!\right)\cos\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right) \\ &\qquad+\sin\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right)\cos\left(\sum_{i=0}^n 2\pi i!\right) \end{aligned}\]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?
JamesJ
  • JamesJ
Yes, but you only have one term after your expansion step: the sin(2.pi.sum from n+1 to infinity ...)
JamesJ
  • JamesJ
...and you dropped a 2pi mid way through but picked it up again; that's a small detail.
KingGeorge
  • KingGeorge
I made a mistake, and had to revise starting from halfway through. I guess that 2pi still slipped through.
KingGeorge
  • KingGeorge
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.
JamesJ
  • JamesJ
Yes
kropot72
  • kropot72
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.
KingGeorge
  • KingGeorge
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.
JamesJ
  • JamesJ
I'll tell you right now the limit is not zero.
KingGeorge
  • KingGeorge
If I had to take a guess, I would say either \(e\) or \(\frac{1}{e}\) or some small multiple of one of them.
JamesJ
  • JamesJ
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.
KingGeorge
  • KingGeorge
Alright. I'll give it some thought. Unfortunately, I've got to now. Hopefully I'll be back later with the full solution.
JamesJ
  • JamesJ
Great.
KingGeorge
  • KingGeorge
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 \[n\sin\left(\frac{2\pi}{n+1}\right)\le n\sin\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right)\le n\sin\left(\frac{2\pi}{n-1}\right).\]
KingGeorge
  • KingGeorge
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 \[\lim_{n\to\infty}n\sin(2\pi en!)=2\pi.\]
JamesJ
  • JamesJ
Yes. The way you wrote the inequality above is slightly confusing, but yep! That's the result. It's very surprising, isn't it?
KingGeorge
  • KingGeorge
Very surprising. Upon first glance it looks very divergent.

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