=+n=1 ln n ns ≥ 2

∞

n=1 ln(2n)

(2n)s − ln 2 2s

∞

n=1 ln n ns ≤ 2

∞

n=1 ln(2n)

(2n)s .

Note that the second inequality in (15.11) can fail when n = 1 and s < 1/ ln 2. In this special case, apply the more delicate inequality ln 1 1s +

ln 2 2s +

ln 3 3s +

ln 4 4s +

ln 5 5s + ≤ 2

ln 2 2s +

ln 4 4s



, which can be demonstrated by showing that g(s) = ln 2 2s − ln 3 3s +

ln 4 4s − ln 5 5s is positive. For this purpose, write g(s) in the equivalent form g(s) = 1

(s − 1)2

 3 2

1 + ln ts−1 ts−1 dt −

 5 4

1 + ln ts−1 ts−1 dt

and prove that the indicated integrand is decreasing in t.)

15.6 Problems 393