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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
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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JamesJBest ResponseYou've already chosen the best response.0
\[ \lim_{n \rightarrow \infty} n.\sin(2\pi.e.n!) \]
 one year ago

amoodaryaBest ResponseYou've already chosen the best response.0
dw:1360717507664:dw
 one year ago

JamesJBest ResponseYou've already chosen the best response.0
Surprisingly, the limit does exist.
 one year ago

kropot72Best ResponseYou've already chosen the best response.0
The maximum value of the sine term is unity. Therefore is the limit infinity?
 one year ago

JamesJBest ResponseYou've already chosen the best response.0
What you've written is not equal to the limit in question.
 one year ago

mykoBest ResponseYou've already chosen the best response.0
just first step towords it... thinking
 one year ago

KingGeorgeBest 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?
 one year ago

JamesJBest ResponseYou've already chosen the best response.0
Yes, but you only have one term after your expansion step: the sin(2.pi.sum from n+1 to infinity ...)
 one year ago

JamesJBest 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.
 one year ago

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

KingGeorgeBest ResponseYou've already chosen the best response.0
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

kropot72Best ResponseYou've already chosen the best response.0
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

KingGeorgeBest ResponseYou've already chosen the best response.0
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

JamesJBest ResponseYou've already chosen the best response.0
I'll tell you right now the limit is not zero.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
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

JamesJBest ResponseYou've already chosen the best response.0
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

KingGeorgeBest ResponseYou've already chosen the best response.0
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

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

KingGeorgeBest ResponseYou've already chosen the best response.0
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

JamesJBest ResponseYou've already chosen the best response.0
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

KingGeorgeBest ResponseYou've already chosen the best response.0
Very surprising. Upon first glance it looks very divergent.
 one year ago
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