Puzzle 11: Solution

Q     Round to the nearest integer:

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 cn(x(1-x)2)n(1-x)99dx/2x, where cn=(3n-2)! / n!(2n-1)! is the yn 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 x(1-x)99dx/2x=(1-x)99dx/2, which is 100/200 = 0.5 exactly. But the sum of cnyn 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 (x-x')(1-x)99dx/2x over x in [1/3,1], where the integrand is always smaller than 3(2/3)99 -- which is tiny. [In fact, since x-x'=0 at x=1/3, the integral turns out to be several orders of magnitude smaller than (2/3)99: approximately 2(2/3)100/1002.] 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/M2; 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 / M2. 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
1/M > 1/(M+1) + 1/(M+2)2 + 3/(M+3)3 + 42/(M+4)4 + 53/(M+5)5 + ... ;
and find approximately the difference when M is great. Hence shew that
1/1001 + 1/10022 + 3/10033 + 42/10044 + 53/10055 + ...
is less than 1/1000 by approximately 10-440.
[I renamed Ramanujan's parameter n as M so that n could be used as a summation index.]
I did not find a solution in the Collected Papers. The following solution should look familiar. Each term nn-2/(M+n)n is the integral of (nn-1/n!)(xe-x)ne-Mxdx/x over the positive real axis. This time nn-1/n! is the zn coefficient of the expansion of x in powers of z=xe-x. So we seem to get the integral of xe-Mxdx/x=e-Mxdx, which equals 1/M exactly; and indeed, if we expand each term nn-2/(M+n)n in a power series in 1/M and sum over n, then we get 1/M. [Direct proof: for k>0, the M-k coefficient is the (k-1)st finite difference of the (k-2)nd degree polynomial nk-2. In the same way it can be seen that the power series for ((t+1)n-2)!(tn+M-1)! / (tn)!((t+1)n+M-1)! sum to 1/tM.] But this shows only that the difference between 1/M and 1/(M+1)+1/(M+2)2+3/(M+3)3+... decreases faster than any power of 1/M. Once more the punchline is the power series in z converges to x only up to the critical point, which here is (x,z)=(1,1/e); for x>1 the series converges to the other, smaller, positive solution x' of z(x')=z(x). So the sum is smaller than 1/M, but only by the integral of (x-x')e-Mxdx/x over x>1, which is asymptotic to 2e-Me/M2 for large M, and in particular is approximately 10-440 for M=1000.

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

nn-2 / (M+n)n = n-2 e-M exp(M2/2n + O(M3/n2)),
so the terms quickly fall below exp(-M) but it takes about eh terms to approximate the sum to within exp(-(M+h)). This behavior, too, is shared by the terms ((t+1)n-2)!(tn+M-1)! / (tn)!((t+1)n+M-1)! in our analogous sums above.