Ramanujan’s Nots by Ramanujan Srinivasa Aiyangar

By Ramanujan Srinivasa Aiyangar

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Example text

49)] and Bailey’s book [4, p. 961. 44) in Hardy’s paper [l]. Example 4. In Corollary 5, let n = 1, replace x by x - 1, and set y = x - 1. Example 5. Zf Re x > ), then 1+3- x-l x+1+5 (x - 1)(x - 2) (x+ I)(x+2)+“‘=x. - 0 10. Hypergeometric Series, 1 21 PR~~E‘. Put n = 1 and replace x by x - 1 in Coroilary 14. 0 Example 5 is given by both Hardy [l, Eq. 41)] and Bailey [4, p. 961. Examlple 6 is also given by Hardy [ 1, Eq. 46)]. Example 6. Zf Re x > 0, then l+- x-l +(x-l)(x-2)+~~~=~-‘T~(x+l)~ ~~ I-(2x + 1) x + 1 (x + 1)(x + 2) PRO~F’.

If x + y + z = 0 and x + y + n is a positive integer, then 3F2[“,=;,,y] r(n + i)r(x + y +~ n + = r(x + n + i)r(y + n + 1) =+y (-XM-Y)k (z),‘k! 1) k=O . PROOF. 2), set a = -x, b = -y, and f = z, and replace n by x + y + 0 n + 1. Entry 34. Zf x and y are arbitrary, then 2Fl(x3Y; Hx Jk-($ + *y + )j + Y + ‘1; 3) = r(LX + +)r(ly 2 + ‘1’ 2 2 Entry 34 is due to Gauss [l]. In Bailey’s text [4, p. 111, Entry 34 is Eq. (2). The following result is due to Kummer [l] and cari be found in Bailey’s monograph [4, p.

3) - k). PROOF. We proceed by induction on r. 1). d = -kil sk+lv(j - 4, lsljlr-1. 4), =- b -k$l kSk+l dr - 4 - kil sk ;gr sj+l dr - k -il) 10. Hypergeometric Series, 1 30 which completes the proof. q Entry 13. Zf Re x > - 1 and r is any positive integer, then f (-XIe k=O (n + k)‘+‘k! 5) = r(n ~ +~ x + 1) cptr). PROOF. Now by Example 5 in Section 10, 2 (-4k k=O (n + k)‘k! WWtx + Os = rtn)rtx + 1) = r(n + x + 1) i r(n + x + 1) CPU). 5) is valid for r = 1. 5) is true with r replaced by r + 1. 5), with r replaced by r + 1, follows.

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