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?
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? 


The generating function of the central binomial coefficients is $$\sum_{n=0}^{\infty}\binom{2n}{n}x^n=\frac{1}{\sqrt{14x}}$$ and so the value of your series is 3. 


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)) 

