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4. 1 is complete. 2 (p. 5). If |b| < 1 and a is an arbitrary complex number, then ∞ ∞ (−1)n (−q; q)n (−aq/b; q)n bn (−1)n (−aq/b; q)n bn q n(n+1)/2 = . 2 (aq; q )n+1 (−b; q)n+1 n=0 n=0 Proof. 1), set h = 2 and t = q 2 , and replace b, c, and a by −b, aq, and aq, respectively. 1) with h = 1, q replaced by q 2 , t = q 2 , and a, b, and c replaced by −b, −bq, and aq 2 , respectively. 1) with q replaced by q 2 , α = −aq/b, β = −bq 2 , and τ = −bq to deduce that, after multiplying both sides by 1/(1 + b), ∞ (−aq/b; q 2 )n (−bq)n 2) (−b; q n+1 n=0 ∞ = (−aq/b; q 2 )n (−aq 2 /b; q 2 )n (−bq 2 )n (−bq)n q 2n (−b; q 2 )n+1 (−bq; q 2 )n+1 n=0 ∞ 2 −2n (1 − aq 4n+2 ) 2 (−aq/b; q 2 )n (−aq 2 /b; q 2 )n b2n q 2n +n (1 − aq 4n+2 ) = (−b; q 2 )n+1 (−bq; q 2 )n+1 n=0 = = ∞ 2 ∞ 2 (−aq/b; q)2n b2n q 2n +n (1 − aq 4n+2 ) (−b; q)2n+2 n=0 (−aq/b; q)2n b2n q 2n (−b; q)2n+1 n=0 ∞ (−aq/b; q)2n b2n q 2n = (−b; q)2n+1 n=0 +n 1+ 2 +n −bq 2n+1 − aq 4n+2 1 + bq 2n+1 ∞ − (−aq/b; q)2n+1 b2n+1 q (n+1)(2n+1) (−b; q)2n+2 n=0 ∞ = (−1)n (−aq/b; q)n bn q n(n+1)/2 , (−b; q)n+1 n=0 which is the desired result.

11). For any complex number a, ∞ (q 2 ; q 4 )∞ ∞ (aq 2 ; q 2 )n q n(n+1)/2 (aq 2 ; q 4 )n q 4n = (aq 4 ; q 4 )∞ (q; q)n (q 2 ; q 2 )2n n=0 n=0 2 ∞ + (aq 2 ; q 4 )∞ 2 (aq 4 ; q 4 )n q 4n +4n+1 . (q 2 ; q 2 )2n+1 n=0 Proof. 13, replace q by q 2 and set b = −1/q. This yields ∞ 1 an q 2n = 4 ; q 4 ) (−q; q 2 ) 2 ; q 4 ) (−q; q 2 ) (q (aq n n ∞ ∞ n=0 + ∞ (aq 2 ; q 4 )n q 4n (q 2 ; q 2 )2n n=0 1 4 4 (aq ; q )∞ (−q; q 2 )∞ ∞ 2 2 (aq 4 ; q 4 )n q 4n +4n+1 . 1). More precisely, let h = 2, c = −q, and a = 0, and let b tend to 0.

17) which is implicit in the work of Ramanujan in his lost notebook [244]. 2]. 10 involving two additional parameters. 10 (p. 10). ∞ 1 qn = 2 (q)n (q)2∞ n=0 ∞ (−1)n q n(n+1)/2 . 18) n=0 Proof. 1), set h = 1, t = c = q, and a = 0, and then let b → 0. 10 follows immediately. 11 (p. 10). ∞ ∞ 1 q 2n = 2 2 (q) (q) n ∞ n=0 (−1)n q n(n+1)/2 1+2 . 19) n=1 Proof. 1), set h = 1, a = 0, c = q, and t = q 2 . Now let b → 0 to deduce that ∞ ∞ n+1 1 q 2n n1 − q q n(n+1)/2 = (1 − q) (−1) 2 2 (q) (q) 1 − q n ∞ n=0 n=0 1 = (q)2∞ = 1 (q)2∞ ∞ ∞ n n(n+1)/2 (−1) q n=0 (−1)n+1 q (n+1)(n+2)/2 + n=0 ∞ (−1)n q n(n+1)/2 1+2 .