Welcome to MathOverflow

Question

1

Whilst trying to solve a combinatorics problem I am faced with summing this series:

1+ 2C_1 2/(3^2) + 4C_2 (2^2)/(3^4) + 6C_3 (2^3)/(3^6)+ ... + 2nC_n (2^n)/(3^(2n))+...

Where 4C_2 is 4 choose 2.

Any idea how to approach this problem?

flag	asked Aug 31 2010 at 11:52 David Willis 63●2

	It's a hypergeometric series; and your sum has a nice closed-form. Try it on Wolfram Alpha. – J. M. Aug 31 2010 at 11:56
	In particular, the series can be turned into \sum_{j=0}^\infty \binom{-1/2}{j}(-4z)^j where z=2/9. – J. M. Aug 31 2010 at 12:03

Gjergji Zaimi · Accepted Answer · 2010-08-31 11:56:44Z UTC

4

The generating function of the central binomial coefficients is $$\sum_{n=0}^{\infty}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$$ and so the value of your series is 3.

answered Aug 31 2010 at 11:56

Gjergji Zaimi
37.8k●2●75●151

Wow, quite close. :) – J. M. Aug 31 2010 at 12:04

Martin Rubey · Answer 2 · 2010-08-31 12:04:23Z UTC

edit: the preceding answer suggests that my browser didn't display the dots, i.e. you really meant the series, not the sequence... Sorry, the below doesn't answer the question.

Does the following look right? (might this be homework?)

(1) -> f n == reduce(+, [binomial(2*i, i)*2^i/3^(2*i) for i in 0..n])
                                                                   Type: Void
(2) -> guess([f n for n in 0..20], maxLevel==2)
   Compiling function f with type NonNegativeInteger -> Fraction(
      Integer)

                  s   - 1
                   21      8p   + 12
                   ++-++     20
                 4  | |    ---------
                    | |    9p   + 18
         n - 1    p  = 0     20
          --+      20
   (2)  [ >      ------------------- + 1]
          --+             9
         s  = 0
          21
                                              Type: List(Expression(Integer))
(3) -> guessPRec [f n for n in 0..20]

   (3)
   [
     [f(n): (9n + 18)f(n + 2) + (- 17n - 30)f(n + 1) + (8n + 12)f(n)= 0,
                     13
      f(0)= 1, f(1)= --]
                      9
     ]
                                              Type: List(Expression(Integer))

In general, it's often a good idea to generalise, i.e., introduce more parameters:

(4) -> f n == reduce(+, [binomial(2*i, i)*x^i/y^(2*i) for i in 0..n])
   Compiled code for f has been cleared.
   1 old definition(s) deleted for function or rule f
                                                                   Type: Void
(5) -> guess([f n for n in 0..20], maxLevel==2)
   Compiling function f with type NonNegativeInteger -> Fraction(
      Polynomial(Integer))

                   s   - 1
                    21      (4p   + 6)x
                    ++-++      20
                 2x  | |    -----------
                     | |              2
         n - 1     p  = 0   (p   + 2)y
          --+       20        20
   (5)  [ >      ---------------------- + 1]
          --+               2
         s  = 0            y
          21
                                              Type: List(Expression(Integer))
(6) -> guessPRec [f n for n in 0..20]

   (6)
   [
     [
       f(n):
                   2                      2
           (n + 2)y f(n + 2) + ((- n - 2)y  + (- 4n - 6)x)f(n + 1)
         +
           (4n + 6)x f(n)
           =
           0
       ,
                      2
                     y  + 2x
      f(0)= 1, f(1)= -------]
                         2
                        y
     ]
                                              Type: List(Expression(Integer))

Trying to sum a series (related to catalan numbers perhaps)

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.