advertisement

Formulas n X n(n + 1) 2 k= k=1 (1) Proof. Add the sum to itself (with the terms in reverse order): 2 nk=1 k = [1 + 2 + . . . + (n − 1) + n] + [n + (n − 1) + . . . + 2 + 1] = [1 + n] + [2 + (n − 1)] + . . . + [(n − 1) + 2] + [n + 1] = n(n + 1). P n X 1 − bn+1 , b 6= 1 b = 1−b k=0 k Pn Proof. Expand the product (1 − b) k=0 ∞ X bk = k=0 Proof. Take the limit of Pn k=0 bk = 1−bn+1 1−b n X Pn bk = k=0 (2) Pn+1 k b bk − k=1 = b0 − bn+1 . 1 , |b| < 1 1−b (3) as n → ∞. ln k = Θ(n ln n) (4) k=0 Proof. First note that the sum isR an approximation sum for an integral: a lower sum for Rn n 1 ln xdx and an upper sum for 2 ln xdx. It follows that Z n ln xdx < 2 n X ln k < Z n ln xdx 1 1 Thus n1 ln k = Θ( 1n ln xdx). Evaluation of the integral by parts yields (n − 1) = Θ(n ln n). P R Rn 1 ln xdx = n ln n − n X 1 = Θ(ln n) k k=1 (5) Proof. As in the argument above, the sum is a lower approximation sum for 1n (1/x)dx Rn Rn and an upper approximation for 2 (1/x)dx, hence the sum is Θ( 1 dx/x). But this integral R is the definition of natural logarithm: 1n dx/x = ln n R ln n! = Θ(n ln n) (6) √ Proof. Applying Sterling’s approximation, we have ln n! = log( 2πn) + n ln(n/e) + ln(1 + Θ(1/n)) = Θ(n ln n). √ n n 1 e n Proof. This is Sterling’s approximation - proof beyond the scope of this page. n! = 2πn 1 1+Θ (7)