In article
<15429522.1215020924777.JavaMail.jakarta@[EMAIL PROTECTED]
>,
Rohan Choudhary <rohan.ckul@[EMAIL PROTECTED]
> wrote:
>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@[EMAIL PROTECTED]
>,
Charles Lazo <laser.lite@[EMAIL PROTECTED]
> wrote:
>>>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.
In article
<23312527.1215030820929.JavaMail.jakarta@[EMAIL PROTECTED]
>,
Rohan Choudhary <rohan.ckul@[EMAIL PROTECTED]
> wrote:
>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@[EMAIL PROTECTED]
>,
Charles Lazo <laser.lite@[EMAIL PROTECTED]
> wrote:
>1/(n+1) + 1/(n+1)(n+2) .... < 1/n + 1/(n^2) + ... = (1/n)(1/(1-1/n)) =
1/(n-1)
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@[EMAIL PROTECTED]
>
take out the trash before replying
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