**A**
Zero -- but only barely. This expression comes to
0.499999999999999999953396...

The n-th term in the sum is the integral from 0 to 1 of
c_{n}(x(1-x)^{2})^{n}(1-x)^{99}dx/2x,
where c_{n}=(3n-2)! / n!(2n-1)!
is the y^{n} coefficient
of the expansion of x in powers of y=x(1-x)^{2}.
Switching sum and integral (which is legitimate since all integrands
are positive), we seem to get 100 times the integral of
^{99}dx/2x=(1-x)^{99}dx/2,_{n}y^{n} converges to x
only up to the critical point 1/3 of y(x). [At that point, y'(x)=0,
and y(x)=2/27 is the radius of convergence of the power series.]
For larger x, the series converges to some x'<x
which is the other positive solution of y(x')=y(x).
Therefore the integral is smaller than the expected 1/200.
The difference, however, is very small: it is the integral of
^{99}dx/2x^{99}^{99}:
approximately 2(2/3)^{100}/100^{2}.]
This explains why our original expression is just below 1/2.
In general, for M>0 the sum of
(3n-2)!(2n+M-1)! / (2n)!(3n+M-1)!
is less than 1/2M, but the difference decreases exponentially
as M grows, and is asymptotic to
2(2/3)^{M}/M^{2};
and likewise for the sum of
((t+1)n-2)!(tn+M-1)! / (tn)!((t+1)n+M-1)!
for any t>0: it is less than 1/tM
by an amount that's asymptotically proportional to
(t / (t+1))^{M} / M^{2}.
The proof is similar, using the power series for the inverse function
of y(1-y)^{t}.

This question is modeled on the following problem by Ramanujan
[Question 526 in the *Journal of the Indian Mathematical Society*
(VI,39), as reported in Ramanujan's *Collected Papers*
(Amer. Math. Society: Providence, 2000), p.329],
which Don Zagier told me some years ago and again at an Oberwolfach
meeting in 2003:

If M is positive shew that[I renamed Ramanujan's parameter n as M so that n could be used as a summation index.]1/M > 1/(M+1) + 1/(M+2) and find approximately the difference when M is great. Hence shew that^{2}+ 3/(M+3)^{3}+ 4^{2}/(M+4)^{4}+ 5^{3}/(M+5)^{5}+ ... ;1/1001 + 1/1002 is less than 1/1000 by approximately 10^{2}+ 3/1003^{3}+ 4^{2}/1004^{4}+ 5^{3}/1005^{5}+ ...^{-440}.

I did not find a solution in the Collected Papers. The following solution should look familiar. Each term n

Don Zagier also notes another devious pitfall in the Ramanujan series: it doesn't converge nearly as fast as the first few terms suggest. For large n we have