Let f(n) be a function on the non-negative integers defined
recursively as follows:

f(0)=1, f(n)=(1+(n*n-n)*f(n-1))/(n*n+1) for n > 0. So f(1)=1/2, f(2)=2/5 ... 

This month's puzzle asks for you to determine the asymptotic behavior
of f(n) as n -> infinity. In other words find a simple function g(x)
such that f(n)/g(n) -> 1 as n -> infinity.

is it cheating to plot the function in excel and look at it?  it
becomes pretty obvious what the asymptotic behavior is...

If we assume that for large n, f(n)~f(n-1) then we can change both
f(n) and f(n-1) to g(n). Solving this equation will give us
g(n)=1/(1+n) and it can be verified that indeed this g is one possible
solution.