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2, p. 169, Entry 7]. A proof via the Poisson summation formula is sketched in [21]. ENTRY 21. 1) Log(-a;q) ro - V_ f ^ — f ? 2) Log f(a,b) = Log(ab;ab) PROOF. Log(-a;q) + I -^ [a. + k=l k(l - a V ) For |q|, |a| < 1, = I Logd + aqn ) = I I {~]) n=0 n=0 k=l r (-1)k"1ak ^ K k=l n=0 f k,n ,{^ ] K ^ (-l)k-]ak k=l k(l - q K ) b } . 32 C. ADIGA, B. C. BERNDT, S. BHARGAVA, AND G. N. 1). ENTRY 22. (i) If | q | < 1, *(q) = f(q,q) = 1 + 2 then ~ (WlJqV^ k2 X q = * j—^— , k=l (q;q^)J-q W L (q2;q2) . -. 3.

186, 187], [23]. Apply the Euler-Maclaurin summation formula [77, p. 128] to the function n(v\ = (m-n) y v x ; - nq = 2 /8(m+n)+mx(x+l)/2+nx(x-l)/2 (2(m+n)x+(m-n)}2/8(m+n) on the interval 1 , x. " 2 9(-'h- -°° < x < °°. The series v B2k (2k-l),( Nu)h2k -°° • k l 1 T2kTT9 where B. denotes the jth Bernoulli number, terminates and, in fact, is J identically equal to zero. Thus, according to Ramanujan, G(q) = op i g(k) k=-°° CHAPTER 16 OF RAMANUJAN'S SECOND NOTEBOOK 39 is perfect and complete, respectively.

WATSON f h1 + /• OO I ^ k=l k even 1 + I a k=l = 2! 1 + + CO I > ak(k+l)/2 k=l k odd k(2k+l) + r ak(2k-l. k=l K K u/ J K K+l y aa k(k-l)/2 ^ " ^ ((aa 3jk(k+l)/2 ) ^ ^^ k=l + y ak(k+l)/2(a3jk(k-l)/2" k=l 2 f(a,a 3x Thirdly, f(-l,a) = I (-i)k(k + l)/2 a k(k-l)/2 + J k=2 k=l upon the replacement of Fourthly, replacing k by M ) k(k-l)/2 a k(k + l)/2 = 0f. k + 1 in the first sum on the right side. (k+n)(k+n-l)/2 ^(n+1)/2 b n(n-l)/2 = a n(n + l)/2 b n(n-l)/2 J a k(k+2n+l)/2 b k(k+2n-l)/2 k=-°° £ {a(ab)n}k(k+1)/2{b(ab)-n}k(k-1>/2 , k=-°° which completes the proof of (iv).