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JamesJ
 3 years ago
Here's a very nice question.
Using the standard power series expansion of e^1, evaluate the following limit ...
JamesJ
 3 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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JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \lim_{n \rightarrow \infty} n.\sin(2\pi.e.n!) \]

amoodarya
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360717507664:dw

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.0Surprisingly, the limit does exist.

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.0The maximum value of the sine term is unity. Therefore is the limit infinity?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360795091206:dw

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.0What you've written is not equal to the limit in question.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just first step towords it... thinking

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0\[\begin{aligned} \sin(2\pi e n!)&=\sin\left(2\pi \left[\sum_{i=0}^\infty \frac{n!}{i!}\right]\right) \\ &=\sin\left(2\pi\left[\sum_{i=0}^ni! +\sum_{j=n+1}^\infty \frac{n!}{j!}\right]\right) \\ &=\sin\left(\sum_{i=0}^n2\pi i!+\sum_{j=n+1}^\infty \frac{n!}{j!}\right) \\ &=\sin\left(\sum_{i=0}^n 2\pi i!\right)\cos\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right) \\ &\qquad+\sin\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right)\cos\left(\sum_{i=0}^n 2\pi i!\right) \end{aligned}\]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
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, but you only have one term after your expansion step: the sin(2.pi.sum from n+1 to infinity ...)

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.0...and you dropped a 2pi mid way through but picked it up again; that's a small detail.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0I made a mistake, and had to revise starting from halfway through. I guess that 2pi still slipped through.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Anyways, 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.

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0I'll tell you right now the limit is not zero.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0If I had to take a guess, I would say either \(e\) or \(\frac{1}{e}\) or some small multiple of one of them.

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.0George, 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
 3 years ago
Best ResponseYou've already chosen the best response.0Alright. I'll give it some thought. Unfortunately, I've got to now. Hopefully I'll be back later with the full solution.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0I think I've finally got it. Notice that for \(n>1\), \[\frac{1}{n+1}\le \sum_{j=n+1}^\infty \frac{n!}{j!}=\frac{1}{(n+1)(n+2)...}\le \sum_{i=0}^\infty \frac{1}{n^i}=\frac{1}{n1}\]So for sufficiently large \(n\), we have \[n\sin\left(\frac{2\pi}{n+1}\right)\le n\sin\left(2\pi\sum_{j=n+1}^\infty \frac{n!}{j!}\right)\le n\sin\left(\frac{2\pi}{n1}\right).\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0It'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
 3 years ago
Best ResponseYou've already chosen the best response.0Yes. The way you wrote the inequality above is slightly confusing, but yep! That's the result. It's very surprising, isn't it?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Very surprising. Upon first glance it looks very divergent.
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