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[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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