On Monday, 4 Dec 2006, 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)
This type of problems can be solved easily but one needs:
1-) A maxima function to test if a function is decreasing.
2-) We have a limit of the type 0 x infinity, infinity with integration
3-) Then if the function beying integrated is decreasing it can be proved that
we can replace the sum with a integration moreover we can replace the function
with another such that limit f/g = 1 (equivalent). A program for doing this
is more or less strait forward to do.
In this case one has the limit = 1/sqrt(n) * integrate(1/sqrt,t,a,inf) = 2
Observe that replacing f by g such that lim f/g = 1 by tailor, this makes the
integration much easier.
1/n * integrate(1/sqrt(n^3+a n^2 + b n + a)) = 1/n * integrate(1/(an^3).
FWIW
Miguel.