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

- JamesJ

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- schrodinger

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- JamesJ

\[ \lim_{n \rightarrow \infty} n.\sin(2\pi.e.n!) \]

- amoodarya

|dw:1360717507664:dw|

- JamesJ

Surprisingly, the limit does exist.

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## More answers

- kropot72

The maximum value of the sine term is unity. Therefore is the limit infinity?

- JamesJ

No

- anonymous

|dw:1360795091206:dw|

- JamesJ

What you've written is not equal to the limit in question.

- anonymous

just first step towords it... thinking

- 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

Yes, but you only have one term after your expansion step: the sin(2.pi.sum from n+1 to infinity ...)

- JamesJ

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

- KingGeorge

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

- 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

Yes

- 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

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

I'll tell you right now the limit is not zero.

- 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

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

Alright. I'll give it some thought. Unfortunately, I've got to now. Hopefully I'll be back later with the full solution.

- JamesJ

Great.

- 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

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

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

Very surprising. Upon first glance it looks very divergent.

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