Re: Limits and continuity

On Wed, 02 Jul 2008 13:48:14 EDT, Rohan Choudhary
<rohan.ckul@[EMAIL PROTECTED]
> wrote in
<news:15429522.1215020924777.JavaMail.jakarta@[EMAIL PROTECTED]
>
in alt.math.undergrad:

> 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