Infinite Series Help
Ravi
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
Ray Vickson
> 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
Robert Israel
> On Jul 9, 12:20=A0pm, Ravi <raram...@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 isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Rob Johnson
Ravi <rara...@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 <r...@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
Ignacio Larrosa Cañestro
Look for [catalan numbers generating function]
--
Best regards,
Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUIT...@mundo-r.com
Rob Johnson
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|