Re: [obm-l] forma fechada e integral

["Cláudio \(Prática\)" <claudio@xxxxxxxxxxxxxxxxxxxxxxx>]

Caro Luís:

Não estou familiarizado com a notação que você usou.

Por acaso, seria isso?

                 infinito
Soma 1 = SOMA   ((n+1)/(2k+1))*C(n+1,2k+1)
                  k = 0

                 infinito
Soma 2 = SOMA  (k/(n+k))*C(n,k)
                 k = 0

                 infinito
Soma 3 = SOMA  (2k/(2n+k))*C(n,k)
                 k = 0

onde:
C(n,k) = no. de subconjuntos de k elementos de um conjunto de n elementos

Um abraço,
Claudio.

----- Original Message -----
From: "Luis Lopes" <llopes@ensrbr.com.br>
To: <obm-l@mat.puc-rio.br>
Sent: Thursday, February 20, 2003 12:55 PM
Subject: [obm-l] forma fechada e integral


> Sauda,c~oes,
>
> \sum_k \frac{n+1}{2k+1} \binom{n+1}{2k+1} .
>
> \sum_{k\geq0} \frac{k}{n+k} \binom{n}{k} .
>
> \sum_{k\geq0} \frac{2k}{2n+k} \binom{n}{k} .
>
> Querendo conhecer as formas fechadas
> (se existentes) das três somas acima,
> escrevi para o prof. Rousseau.
>
> Em função das suas respostas, fiquei sabendo
> que não existem. Mas não entendi a passagem
> para a integral e a justificativa decorrente.
> Para não incomodá-lo MAIS uma vez, gostaria
> de perguntar antes pra lista (e participar também
> tais resultados). Talvez a resposta até seja
> elementar.
>
> Cortando algumas partes, aí segue nossa
> discussão.
>
> []'s
> Luís
>
>
> Dear Cecil,
>
> Retaking my CRUX saga, consider problem
> 2683 whose solution appears in 28(8),
> December 2002, pp~539--540.
>
> Find the value of \lim_{n\to\infty} \left(
> \frac{1}{2^n} \sum_{k=0}^{\lfloor n/2\rfloor}
> \frac{n+1}{2k+1} \binom{n+1}{2k+1} \right) .
> It turns out to be 2.
>
> In this problem - as always - I am more
> interested in a closed form to
>
> \sum_k \frac{n+1}{2k+1} \binom{n+1}{2k+1} .
>
> As CRUX didn´t mention it, I strongly suspect
> there is none.
>
> Regards,
> Luis
>
> Dear Luis:
>
> I would be very surprised if there is
> a closed form for this.  One can write
> it rather compactly as an integral, but
> that doesn't seem to help very much.
> You can certainly put it in hypergeometric
> form, but not as far as I know can it
> be written in a form where the sum can
> be deduced from one of the classical
> formulas (Gauss. Dixon, Pfaff-Saalschutz)
>
> Cecil
>
>
> %%%%% Segunda mensagem%%%%
> Luis Lopes wrote:
>
> Dear Cecil,
>
> I knew already the published solution
> (similar to yours). .......
>
> Again there shouldn´t be any closed form to
> \sum_{k\geq0} \frac{k}{n+k} \binom{n}{k} .
>
> Not to mention
>
> \sum_{k\geq0} \frac{2k}{2n+k} \binom{n}{k}
>
> Thank you,
> Luis
>
> Dear Luis:
>
> The first one is the same as  n \int_0^1 (1+x)^{n-1} x^n dx
> and I am pretty sure there is no closed form for the integral.
> The second one is similar.  You could get something for it
> if you could evaluate \int_0^1 (1+x)^{n-1} x^{2n}  dx,
> and I believe that this is out of reach.  I haven't tried that
> hard, but Maple fails to give an evaluation and nothing
> I have found in Gradshteyn and Ryzhik is helpful.
>
> Cecil
>
>
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