Sum of near infinite series?

Having some trouble figuring out where to start with this problem:
Find the product of the first n (n=9999) terms of the following expression where n is an integer variable greater than one:
eq = 2/1 * 2/3 * 4/3 * 4/5 * 6/5 * 6/7 * 8/7 * 8/9 .... *n-1/n
Any help on figuring this out would be appreciated.

Answer 1

[David Hill]

p=[];
for i=1:2:n+1
    p=[p,(n+1)/n,(n+1)/(n+2)];
end
P=prod(p(1:n));

[John Mitchell]

Thank you for responding, but I forgot to mention that we can't use for loops in the problem, is there any way to do them without loops?

[David Hill]

a=2*[1:floor((n+1)/2);1:floor((n+1)/2)];
nu=a(:)';
d=[1,nu-1];
p=prod(nu(1:n)/d(1:n));

Answer 2

[Stephan]

Think about your row. You can rewrite it - always two elements are the same as n^2/(n^2-1).

With this knowledge read about cumprod. This should be enough to bring you to work the rest by yourself.