Concrete mathematics : a foundation for computer science

Why isn’t it
r(k) = k + 1 ?
Oh, I see.
5.7 PARTIAL HYPERGEOMETRIC SUMS 227
Time for an example. Let’s try the partial sum (5.114); Gosper’s method
should be able to deduce the value of
for any fixed n. Ignoring factors that don’t involve k, we want the sum of
The first step is to put the term ratio into the required form (5.117); we have
~
t(k+ 1)
= (k-n) P(k+ 1) q(k)
t(k) ~ (k+ 1) = p(k)r(k+ 1)
so we simply take p(k) = 1, q(k) = k - n, and r(k) = k. This choice of
p, q, and r satisfies (5.118), unless n is a negative integer; let’s suppose it
isn’t. According to (5.1~3)~ we should consider the polynomials Q(k) = -n
and R(k) = 2k - n. Since R has larger degree than Q, we need to look at
two cases. Either d = deg(p) - deg(R) + 1, which is 0; or d = -26/y where
(3 = -n and y = 2, hence d = n. The first case is nicer, so let’s try it first:
Equation (5.121) is
1 = (k-n)cxc-k%
and so we choose 0~0 = -l/n. This satisfies the required conditions and gives
CT(k) =
r(k) s(k) t(k)
p(k)
-,(li n
~ .-. k (-l)k
n 0
n - l
k-, (-W’ 9
=( >
which is the answer we were hoping to confirm.
If we apply the same method to find the indefinite sum 1 (z) 6k, without
the (-1 )k, everything will be almost the same except that q(k) will be n - k;
hence Q(k) = n - 2k will have greater degree than R(k) = n, and we will
conclude that d has the impossible value deg(p)’ - deg(Q) = -1. Therefore
the function (c) is not summable in hypergeometric terms.
However, once we have eliminated the impossible, whatever remainshowever
improbable-must be the truth (according to S. Holmes [70]). When
we defined p, q, and r we decided to ignore the possibility that n might be a