By Maring S
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Extra resources for Oracle9i CORBA Developer's Guide and Reference (Part No. A90187-01) (Release 9.0.1)
Example text
K=1 n JO (32a) (2k − 1)3 = n2 (2n2 − 1) 3. 124 q k n2 − k 2 = 1. 10 k=1 1 q(q + 1) 2n2 − q 2 − q 4 [q = 1, 2, . ] 1 n(n + 1) 12n3 + 63n2 + 107n + 58 60 n k! · k = (n + 1)! 126 k=0 (n + k)! (n − k)! 13111 k=1 1 1 = C + ln n + − k 2n ∞ k=2 Ak , n(n + 1) . . 133 k=2 23 − 1 B4 1 B2 1 = (C + ln n) + ln 2 + 2 + + ... 14 Sums of products of reciprocals of natural numbers n 1. k=1 n 2. k=1 n 1 = [p + (k − 1)q](p + kq) p(p + nq) GI III (64)a n(2p + nq + q) 1 = [p + (k − 1)q](p + kq)[p + (k + 1)q] 2p(p + q)(p + nq)[p + (n + 1)q] GI III (65)a n 3.
1 = ln 2 k (−1)k+1 1 =1−2 2k − 1 ∞ k=1 3. ∗ ∞ k=1 rq a + 1 − q (1 − q)2 (−1)k+1 k=1 2. [|q| < 1] ka 1 = bk (b − 1)a+1 [|q| < 1] (cf. 113) (cf. 511) a ∞ k=1 π 1 = (4k − 1)(4k + 1) 4 ⎡ ⎣ 1 ba−i i=1 i j=0 (cf. (a + 1 − j)! [a = 1, 2, 3, . . 233 ∞ 1. k=1 1 1 1 = 1 + p + p + . . = ζ(p) kp 2 3 ∞ (−1)k+1 2. 10 k=1 1 22n−1 π 2n |B2n |, = k 2n (2n)! ∞ (−1)k+1 4. k=1 ∞ 5. k=1 (−1)k+1 k=1 ∞ k=1 1 π2 = k2 6 1 (22n−1 − 1)π 2n |B2n | = k 2n (2n)! 1 (22n − 1)π 2n |B2n | = 2n (2k − 1) 2 · (2n)! ∞ 6. 1 = (1 − 21−p ) ζ(p) kp 1 π 2n+1 |E2n | = 2n+2 2n+1 (2k − 1) 2 (2n)!
K=n+1 n (−1)r 2. 211 uk = u1 + u2 + u3 + . . 212 k=1 composed of the absolute values of its terms converges. 211 is said to converge conditionally. Every absolutely convergent series converges. 211 converges absolutely. 211 diverges. 211 converges absolutely. 211 diverges. 211 diverges. 211 converges absolutely. 211 diverges. 224 Suppose that f (x) is a positive decreasing function and that ek f ek lim =q k→∞ f (k) ∞ for natural k. If q < 1, the series k=1 f (k) converges. If q > 1, this series diverges.
Oracle9i CORBA Developer's Guide and Reference (Part No. A90187-01) (Release 9.0.1) by Maring S
by Edward
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