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From: Rob Johnson on 9 Jul 2010 18:27
In article <20100709.141148(a)whim.org>,
Rob Johnson <rob(a)trash.whim.org> wrote:
>In article <c781a3a3-2633-42ff-a2e2-9ace9abd3bd4(a)5g2000yqz.googlegroups.com>,
>Ravi <raramasw(a)gmail.com> wrote:
>>Does this infinite series have a closed form?
>>
>>Sum { from k = 1 to inf, (2k choose k+1) [p (1-p)]^k }.
>
>First, a couple of identities. The first follows from the binomial
>theorem:
>
> C(-1/2,k) = C(2k,k)(-1/4)^k [1]
>
>The second follows from the recursive relation defining Pascal's
>Triangle:
>
> C(2k,k+1) = C(2k+1,k+1) - C(2k,k)
>
> = 1/2 C(2k+2,k+1) - C(2k,k) [2]
>
>Now, to the sum in question. Let x = p(1-p), then
>
> oo
> --- k
> > C(2k,k+1) x
> ---
> k=1
>
> oo
> --- k
> = > C(2k,k+1) x
> ---
> k=0
>
> oo
> --- k
> = > (1/2 C(2k+2,k+1) - C(2k,k)) x
> ---
> k=0
>
> oo
> --- k+1 k k
> = > (1/2 C(-1/2,k+1)(-4) - C(-1/2,k)(-4) ) x
> ---
> k=0
>
> 1 1 1
> = --- ( ---------- - 1 ) - ( ---------- )
> 2 x sqrt(1-4x) sqrt(1-4x)
>
> 2 1
> = ----------------- - ---------- [3]
> 1-4x + sqrt(1-4x) sqrt(1-4x)
>
>To get the final formula, substitute x = p(1-p).

In comparing my result with Robert Israel's, I see that

1-4x = 1-4p(1-p) = (1-2p)^2

Thus, [3] can be simplified to

oo
--- k
> C(2k,k+1) [p(1-p)]
---
k=1

2 1
= ----------------- - ------
(1-2p)^2 + |1-2p| |1-2p|

1 1 - |1-2p|
= ------ ---------- [4]
|1-2p| 1 + |1-2p|

Rob Johnson <rob(a)trash.whim.org>
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