Question

The question is in the comment. Be patient, I am a slow typist.

Answer 1

Prove that
\[\Large
\lim_{n\to \infty } \sin(n!)
\]
does not exist.
Here is a proof that
\[\Large
\lim_{n\to \infty } \sin(n)
\]
does not exist.
Suppose not and let
\[
\Large
a=\lim_{n\to \infty } \sin(n)
\]
Case 1: \( a\ne 0\)

Write
\[
\sin(n+1) -\sin(n+1) = 2 \sin(n) \cos(1)\\
\lim_{n\to \infty }(\sin(n+1) -\sin(n+1)) = \lim_{n\to \infty }( 2 \sin(n) \cos(1)\\)\\
a-a= 2 a \cos(1)\\
\cos(1)=0

\]
contradiction.
case 2. a=0
Write
\[
\sin(n+1) =\sin(n)cos(1)+ \cos(n) \sin(1)\\
\lim_{n\to \infty }\sin(n+1) =\lim_{n\to \infty }\sin(n)cos(1)+ \lim_{n\to \infty }\cos(n) \sin(1)\\
0=0 +\lim_{n\to \infty }\cos(n) \sin(1)\\so\\
\lim_{n\to \infty }\cos(n)=0
\]
This is a contradiction since
\[
\sin^2(n) + \cos^2(n)=1
\]

Answer 2

@eliassaab: For case 1, I presume there is a typo in the following:

sin(n+1) - sin(n+1) = 0 [not 2sin(n)cos(1)].

Please advise. Thanks.

Answer 3

Sorry
sin(n+1) - sin(n-1) = 2sin(n)cos(1)

Answer 4

I think you meant the following:

sin(n+1) - sin(1-n) = 2sin(n)cos(1)?

Please advise. Thanks.

Answer 5

Yes

Answer 6

Thanks but this does not affect the rest of the argument.

Answer 7

Here a small correction of my first post. Thanks for @shiraz14 \[
\sin (n + 1) - \sin (1 - n) = 2\sin (n)\cos (1) \\
\lim_ {n\to\infty} (\sin (n + 1) - \sin (1 - n)) = \lim_ {n\to
\infty} (2\sin (n)\cos (1) \\) \\
a +a = 2 a\cos (1) \\
\cos (1) = 1
\]

Answer 8

@eliassaab: Your last post is correct. And yes, this does raise an important (& interesting) point which I spent some time clarifying & debating about (in a forum elsewhere in the past) although no conclusion was ever reached (due to its abstraction) - is infinity+1=infinity? And what about infinty^2?

Answer 9

Where did I use (due to its abstraction) - is infinity+1=infinity? And what about infinty^2? @shiraz14

Answer 10

@eliassaab : No, this was addressed in a different debate, but its implications are widespread, including your proof above. Perhaps we can discuss this in a different thread.

All the Best.

Answer 11

We still need to show (notice the n!) nafactorial
\[
\Large
\lim_{n\to \infty } \sin(n!)
\]
does not exist.
@experimentX @ganeshie8 @ranga @shiras14

Answer 12

@shiraz14

Answer 13

will this work @eliassaab ?

say, \(t= n!\)
as \(n \to \infty\), \(t \to \infty\)

\[

\lim_{t\to \infty } \sin(t) = DNE
\]

Answer 14

You know that if a sub-sequence converges, it does not mean that the sequence does converge in general.

Answer 15

I see the error...

Answer 16

Interestingly, \[\lim_{n\to\infty}\sin(2\pi en!)\]converges. However, I can't see any easy way to show that \(\lim_{n\to\infty}\sin(n!)\) converges. In fact, after searching the internet for a while, the only thing I could come up with, was that the convergence of \(\lim_{n\to\infty}\cos(n!)\) was actually an open problem.

Answer 17

I conjecture that it diverges and I did not know that it is an open problem.
http://www.wolframalpha.com/input/?i=list+plot++the+sequence+sin%28n%21%29

Answer 18

Judging from the graphs that wolfram is giving I too would guess that it diverges, but I really don't know of a good way to approach the problem. In the proof that I've seen of the convergence of \(\sin(2\pi en!)\) it depends heavily on both the \(2\pi\), and the series expansion of \(e\).

Answer 19

This is about the most useful thing I was able to scrounge up from the internet.
http://math.stackexchange.com/questions/8690/is-there-a-limit-of-cos-n
and this is related
http://mathoverflow.net/questions/45665/distribution-mod-1-of-factorial-multiples-of-real-numbers