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" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> 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 > > > ========================================================================= > Instruções para entrar na lista, sair da lista e usar a lista em > http://www.mat.puc-rio.br/~nicolau/olimp/obm-l.html > O administrador desta lista é <[EMAIL PROTECTED]> > ========================================================================= ========================================================================= Instruções para entrar na lista, sair da lista e usar a lista em http://www.mat.puc-rio.br/~nicolau/olimp/obm-l.html O administrador desta lista é <[EMAIL PROTECTED]> =========================================================================