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From: Ravi on 9 Jul 2010 15:20
Hi -

Does this infinite series have a closed form?

Sum { from k = 1 to inf, (2k choose k+1) [p (1-p)]^k }.

Apologies for the "non-math" format - hope you can understand it

Thanks

Ravi
From: Ray Vickson on 9 Jul 2010 16:09
On Jul 9, 12:20 pm, Ravi <raram...(a)gmail.com> wrote:
> Hi -
>
> Does this infinite series have a closed form?
>
> Sum { from k = 1 to inf, (2k choose k+1) [p (1-p)]^k }.

This = p*sum(C(2k,k)*(1-p)^k,k=1..infinity). According to Maple 9.5:
S:=p*sum(binomial(2*k,k+1)*(1-p)^k,k=1..infinity);

4 p (-1 + p)
S := - ----------------------------------
1/2 1/2 2
(-3 + 4 p) (1 + (-3 + 4 p) )

R.G. Vickson

>
> Apologies for the "non-math" format - hope you can understand it
>
> Thanks
>
> Ravi


From: Robert Israel on 9 Jul 2010 16:21
Ray Vickson <RGVickson(a)shaw.ca> writes:

> On Jul 9, 12:20=A0pm, Ravi <raram...(a)gmail.com> wrote:
> > Hi -
> >
> > Does this infinite series have a closed form?
> >
> > Sum { from k =3D 1 to inf, (2k choose k+1) [p (1-p)]^k }.
>
> This =3D p*sum(C(2k,k)*(1-p)^k,k=3D1..infinity). According to Maple 9.5:
> S:=3Dp*sum(binomial(2*k,k+1)*(1-p)^k,k=3D1..infinity);
>
> 4 p (-1 + p)
> S :=3D - ----------------------------------
> 1/2 1/2 2
> (-3 + 4 p) (1 + (-3 + 4 p) )
>

.... except that you have 1-p where the question had p*(1-p).
So it should be

S := sum(binomial(2*n,n+1)*(p*(1-p))^n, n=1..infinity);

4 p (p - 1)
- --------------------------------------------
2
(1/2) / (1/2)\
/ 2\ | / 2\ |
\(-1 + 2 p) / \1 + \(-1 + 2 p) / /

And this is only valid for |p(p-1)| < 1/4 (in particular, not for p=1/2).
If 0 < p < 1/2 it simplifies to p/((1-p)(1-2p)).
If 1/2 < p < 1 it simplifies to (1-p)/(p (2p - 1)).
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Rob Johnson on 9 Jul 2010 18:01
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).

Rob Johnson <rob(a)trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
From: Ignacio Larrosa Cañestro on 9 Jul 2010 18:13
Ravi wrote:
> Hi -
>
> Does this infinite series have a closed form?
>
> Sum { from k = 1 to inf, (2k choose k+1) [p (1-p)]^k }.
>
> Apologies for the "non-math" format - hope you can understand it
>
> Thanks
>
> Ravi

Look for [catalan numbers generating function]


--
Best regards,

Ignacio Larrosa Ca�estro
A Coru�a (Espa�a)
ilarrosaQUITARMAYUSCULAS(a)mundo-r.com

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