On Mon, Dec 04, 2006 at 01:01:59PM -0800, Daniel Lakeland wrote:
> On Sun, Dec 03, 2006 at 08:49:50PM +0000, Leo wrote:
> > Hi all,
> >
> > I am learning maxima. I want to use it to get the value of the
> > following when n goes to infinity.
> >
> > n
> > ====
> > \ 1
> > > -----------
> > / sqrt(n + i)
> > ====
> > i = 1
> > -----------------
> > sqrt(n)
>
> You could compare it to the following integral:
>
> integrate(1/sqrt(n) * 1/(sqrt(n+i)),i,0,n);
>
> maxima will calculate this and take the limit as n -> inf
>
> Perhaps this will help you.
Some further help: the floor function can be used to visualize the sum
as the area under a curve. For example, when n = 20
plot2d(ev(1/sqrt(n) * 1/sqrt(n+floor(i)),n=20),[i,1,20]);
you can compare this to the plot of both
1/sqrt(n) * 1/sqrt(n+i)
as well as
1/sqrt(n) * 1/sqrt(n+i+1)
These two integrals "sandwich" the sum you're looking for.
If you can prove that the two integrals have the same limit as n ->
infinity then the sum must have the same limit. Maxima will do this
for you. Give it a try.
In general this is an example of how maxima is not clever enough to
solve the problem, but it's very good at helping you be clever.
--
Daniel Lakeland
dlakelan at street-artists.org
http://www.street-artists.org/~dlakelan