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

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

Mathematics
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\[ \lim_{n \rightarrow \infty} n.\sin(2\pi.e.n!) \]
|dw:1360717507664:dw|
Surprisingly, the limit does exist.

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

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