1550251515-Classical_Complex_Analysis__Gonzalez_

Sequences, Series, and Special Functions 551

show that the Bernoulli numbers also satisfy the following symbolic

equations.

(a) Bn - (B - 1r = (-1r-^1 n

(b) (B + l)n -(B -1r = (-1r-^1 n (n:::: 2)

11. Let f(z) be an entire function. From (8.3-6) we have, letting h = Bz,
    then h = (B + l)z,

f(a + Bz) = f(a) + J'(a)Bz + f"~a) B^2 z^2 + ...

2.

f (a+ (B + l)z) = J(a) + J'(a)(B + l)z + f'~~a) (B + 1)^2 z^2 + · ·.

so that

(1) f(a + (B + l)z)-J(a + Bz) = J'(a)z

since (B + l)n - Bn = 0 for n :2: 2. In particular, ford= 0 we have

J((B + l)z) - f(Bz) = f'(O)z

Now, in formula (1) put z = 1 and let a= 1, 2, ... , n to show that

(2) f'(l) + !'(2) · · · + J'(n) = J(B + 1 + n) - f(B + 1)

12. Apply formula (2) of problem 11 to the function f(z) = zP+l to obtain
    (assuming p a positive integer) the following formula:
    nP+l p
    8 =1P+2P+···+nP = --+nP(B1+l)+ -nP-^1 B2

p p+ 1 2!

and show that

+ p(p -^1 ) nP-^2 B3 + .. · + nB

3! p

81 =^1 / 2 n(n + 1)

82 =^1 / 6 n(n + 1)(2n + 1)

83 = l/4n2(n + 1)2

84=1fa 0 n(n + 1)(2n + 1)(3n^2 + 3n - 1)

*13. Let f(z) = :L:::'=o anzn for lzl < R, and let

M(r) =max lf(z)I

lzl=r

where 0 < r < R. Use Cauchy's inequality (7.23-1) to show that

(n = 0, 1, 2, ... )