By Mitsuyasu Hashimoto
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2n (−2n)! n! (−n − 12 )! (− 12 )! This is a rather distressing development. We were expecting an answer in simple form, but the answer that we are looking at contains some factorials of negative numbers, 46 The Hypergeometric Database and some of these negative numbers are negative integers, which are precisely the places where the factorial function is undefined. Fortunately, what we have is a ratio of two factorials at negative integers; if we take an appropriate limit, the singularities will cancel, and a pleasant limiting ratio will ensue.
2. ). To identify this series, note that the smallest value of k for which the term tk is nonzero is the term with k = −1. Hence we begin by shifting the origin of the sum as follows: 1 1 = . (2k + 1)(2k + 3)! k≥0 (2k − 1)(2k + 1)! k≥−1 The ratio of two consecutive terms is (k − 12 ) tk+1 1 = . 1) Hence our given series is identified as 1 − 12 = − 1F2 1 (2k + 1)(2k + 3)! 2 k 3 2 ; 1 . 3. Suppose we define the symbol [x, d]n = n−1 j=0 (x 1, − jd), if n > 0; if n = 0. Now consider the series k n [x, d]k [y, d]n−k .
First cancel out all factors that look like cn or ck (in this case, a factor of 2−n ) that can be cancelled. Then replace every binomial coefficient in sight by the quotient of factorials that it represents. Finally, cancel out all of the factorials by suitable divisions, leaving only a polynomial identity that involves n and k. ” In this case, after multiplying through by 2n , and replacing all of the binomial coefficients by their factorial forms, we obtain (n + 1)! n! n! n! − =− + 2k! (n + 1 − k)!