644 Series
involving binomial coefficients. For example, putting s = 3 gives
n
k3 = n3 *[n4(1) + 4n3(-C + 6n2(+) + 4n(0)],
k=1
so
Putting s = 4 gives
so
n(n+1)(2n+1)(3t2+3n-1)
30
Some people spend huge amounts of time working out these formulas by other
methods.
3 Show that setting p = 1, q = n, f(x) = x-1, m = 3, and C1 = y in (12.665)
gives the formula
(1)
4
k3 =n2(n + 1)2
k1^4
n
0 = n4 + 3[ns + 5n4(- ) + 10n3(g) + 10n2(0) +
k=1
V -n(6n4 + 15n3 + iOn2 - 1)
k=1^30
Ik=logn+y+2n-12n2+6
k l
where y is the Euler constant. Show that setting p = 1, q = 10, f(x) = z 1,
and C1 = y in (12.666) gives the formula
B,-J10
X I
(2) y kIlk - log 10- + 4 10 m+1B,, x)dx.
7'2 1
Remark: Knowing that
log 10 = 2.30258 50929 94046
it is not difficult to push a pencil through the calculations by which (2) is used
with m = 10 to obtain the first 10 decimal places in
y = 0.57721 56649 01532 86061
When y is known, (1) enables us to make very close estimates of the sum of
10 or more terms of the harmonic series.
4 Supposing that z > -1, show that putting p = 1, q = n, and f(x)
log (z + x) in (12.663) gives, when m z 2,
n
(1) E log(z+k)=(z+n+y)log(z+n)-(z+-1T)log(z+1)-n+1
k-1
+ m
B,
L
1 - 1 +
I
n (m - 1)!Bm(x)dx.
,1-z (9 - 1)1 (z - (z 1),-1^1 (z + x)"