A ANSWERS TO EXERCISES 517
by Vandermonde’s convolution (5.92). 5.51 (a) Reflection gives F(a, -n; 2a; 2) = (-1 )“F( a, -n; 2a; 2). (Incidentally, this formula implies the remarkable identity A2”‘+’ f(0) = 0, when f(n) = 2nxc/(2x)“.>~ (b) The term-by-term limit is &kSm (r) m(-2)k plus an additional term for k = 2m - 1: the additional term is (-m)... (-1) (1)...(m) (-2m+ 1) . . . (-1)22m+’ I:-2m). ( - 1 ) ( 2 m - l)! ,I ,I pm+1 -2 = (-ltm+'* =-
(CL') '
hence, by (5.104), this limit is -l/( y2), the negative of what we had. 5.52 The terms of both series are zero for k > N. This identity corresponds to replacing k by N - k. Notice that
When b = -i, the left side of (5.110) is 1 - 22 and the right side is (1 -42+422)"2, independent of a. The right side is the formal power series
5.53
l/2 l+ ( 1 )
42(2-l)+
l/2 (
2
1
16z2(z-1)2+~~~,
which can be expanded and rearranged to give 1 - 22+ Oz2 + Oz3 f. ; but the rearrangement involves divergent series in its intermediate steps when z = 1, so it is not legitimate. 5.54
If m + n is odd, say 2N - 1, we want to show that lim F E'O
N-m-;, -N+c 1 =o. -m+e ( 1)
Equation (5.92) applies, since -m + c > -m - i + E, and the denominator factor T(c-b) = T(N-m) is infinite since N < m; the other factors are finite. Otherwise m + n is even; setting n = m ~ 2N we have fi,mo F
(
-N, N-m-i+e
1
-m+c
1)
=
(N-1/21N rnN
by (5.93). The remaining job is to show that (N - l/2)! (m-N)!
-(-l/2)!
m!
=
