* [2006.12.10] Miguel Lopez wrote:
^^^^^^^^^^^^
> 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.
I don't quite understand this. Could you try to use it to solve the
original limit problem? I have another approach leading to 2*sqrt(2)-2
but I am not 100% sure.
--
Leo