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Rohan Choudhary
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 Posted: Wed Jul 02, 2008 5:48 pm Post subject: Limits and continuity |
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Can someone help me with this :
Lim(n->inf) nSin(2(pi)(e)n!)
I believe that limit does not exist. One of the guys tried to prove that the limit is 2(pi) by expanding e. |
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Charles Lazo
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| Posted: Wed Jul 02, 2008 7:49 pm Post subject: Re: Limits and continuity |
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One of the guys tried to prove that the limit is 2(pi) by expanding e.
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Rohan, he was right. Let (e)n! = I + f, where I is the integer part and f is the fraction. Then sin(2(pi)(e)n!) = sin(2(pi)f). f = 1/(n+1) + 1/(n+1)(n+2) + ... So sin(2(pi)(e)n!) may be approximated by sin(2(pi)/(n+1)) which tends to 2(pi)/(n+1) in the limit.
Namaste,
-- Charles |
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Rohan Choudhary
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| Posted: Wed Jul 02, 2008 8:33 pm Post subject: Re: Limits and continuity |
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My only doubt in this is that how do you prove that 1/(n+1) + 1/(n+1)(n+2) .... where this sequence itself contains infinite terms and has each term tending to 0, how is it that its limit is 0, because thats what the whole thing depends upon.
If someone can do that more rigorously, it wud be great.
Thanks for the help. |
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Charles Lazo
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| Posted: Wed Jul 02, 2008 9:16 pm Post subject: Re: Limits and continuity |
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1/(n+1) + 1/(n+1)(n+2) .... < 1/n + 1/(n^2) + ... = (1/n)(1/(1-1/n)) = 1/(n-1)
-- Charles |
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Brian M. Scott
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| Posted: Thu Jul 03, 2008 12:39 am Post subject: Re: Limits and continuity |
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On Wed, 02 Jul 2008 13:48:14 EDT, Rohan Choudhary
<rohan.ckul@gmail.com> wrote in
<news:15429522.1215020924777.JavaMail.jakarta@nitrogen.mathforum.org>
in alt.math.undergrad:
| Quote: |
Can someone help me with this :
Lim(n->inf) nSin(2(pi)(e)n!)
I believe that limit does not exist. One of the guys tried
to prove that the limit is 2(pi) by expanding e.
|
He's correct, and his approach can be made to work.
e = sum_{k >= 0}{1/k!}, so for a fixed n,
e * n! = n! * sum_{k >= 0}{1/k!} =
sum_{k = 0}^n {n!/k!} + sum_{k > n} {n!/k!}.
Let m(n) = sum_{k = 0}^n {n!/k!}; clearly m(n) is an
integer, and hence
sin(2 * pi * e * n!) =
sin(2 * pi * m(n) + 2 * pi * sum_{k > n} {n!/k!}) =
sin(2 * pi * sum_{k > n} {n!/k!}) =
sin(2 * pi * (1/(n+1) + 1/[(n+1)(n+2)] + ...)).
Now you need to show that
n * sin(2 * pi * (1/(n+1) + 1/[(n+1)(n+2)] + ...))
approaches 2 * pi as n increases. Hint: You'll want to use
the fact that sin(x)/x --> 1 as x --> 0.
Brian |
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Rob Johnson
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| Posted: Thu Jul 03, 2008 4:36 pm Post subject: Re: Limits and continuity |
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In article <15429522.1215020924777.JavaMail.jakarta@nitrogen.mathforum.org>,
Rohan Choudhary <rohan.ckul@gmail.com> wrote:
| Quote: |
Can someone help me with this :
Lim(n->inf) nSin(2(pi)(e)n!)
I believe that limit does not exist. One of the guys tried to prove that the limit is 2(pi) by expanding e.
|
In article <30429294.1215028220677.JavaMail.jakarta@nitrogen.mathforum.org>,
Charles Lazo <laser.lite@hotmail.com> wrote:
| Quote: |
One of the guys tried to prove that the limit is 2(pi) by expanding e.
Rohan, he was right. Let (e)n! = I + f, where I is the integer part and f is the fraction. Then sin(2(pi)(e)n!) = sin(2(pi)f). f = 1/(n+1) + 1/(n+1)(n+2) + ... So sin(2(pi)(e)n!) may be approximated by sin(2(pi)/(n+1)) which tends to 2(pi)/(n+1) in the limit.
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In article <23312527.1215030820929.JavaMail.jakarta@nitrogen.mathforum.org>,
Rohan Choudhary <rohan.ckul@gmail.com> wrote:
| Quote: |
My only doubt in this is that how do you prove that 1/(n+1) + 1/(n+1)(n+2) .... where this sequence itself contains infinite terms and has each term tending to 0, how is it that its limit is 0, because thats what the whole thing depends upon.
If someone can do that more rigorously, it wud be great.
|
In article <30159206.1215033392801.JavaMail.jakarta@nitrogen.mathforum.org>,
Charles Lazo <laser.lite@hotmail.com> wrote:
| Quote: |
1/(n+1) + 1/(n+1)(n+2) .... < 1/n + 1/(n^2) + ... = (1/n)(1/(1-1/n)) = 1/(n-1)
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It makes reading your reply so much easier if you quote the article
to which you are replying.
Why not just
1
---
n+1
1 1 1 1 1 1 1 1 1 1
< --- + --- --- + --- --- --- + --- --- --- --- + ...
n+1 n+1 n+2 n+1 n+2 n+3 n+1 n+2 n+3 n+4
1 1 1 1 1 1 1 1 1 1
< --- + --- --- + --- --- --- + --- --- --- --- + ...
n+1 n+1 n+1 n+1 n+1 n+1 n+1 n+1 n+1 n+1
1
= -
n
The Sandwich Theorem <http://en.wikipedia.org/wiki/Squeeze_theorem>
says that
2pi 2pi
lim n sin( --- ) <= lim n sin(2pi e n!) <= lim n sin( --- )
n->oo n+1 n->oo n->oo n
Using lim_{x->0} sin(x)/x = 1, we get
2pi n n+1 2pi
lim n sin( --- ) = 2pi lim --- --- sin( --- ) = 2 pi
n->oo n+1 n->oo n+1 2pi n+1
and
2pi n 2pi
lim n sin( --- ) = 2pi lim --- sin( --- ) = 2 pi
n->oo n n->oo 2pi n
Therefore
lim n sin(2pi e n!) = 2 pi
n->oo
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font |
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