Calculus: Analytic Geometry and Calculus, with Vectors

602 Series

way to appraise the sum of the areas of the triangular patches is to put

duplicates of these patches in the rectangle having opposite vertices at

the origin and the point (l,ui). Setting

,4n=IT1I+IT2I+ +ITn-iI,

we see that 0 < 142 < 143 <. < 4 < it,. There is therefore a

number C such that

0 < lim [IT,I+IT2I+ +ITn_iIl =C<ul.

n-.w

Putting C. = A + is,, gives the following theorem.

Theorem 12.25 (integral test) If f is positive and continuous and

decreasing over the interval x > 1, if f(x) -+ 0 as x -). -, and if uk = f(k)

for each k = 1, 2, 3, ,then the sequence C1, C2, C3, of constants

defined by

n

z1

uk =fif(x) dx + C

is convergent and 0 < Cn < u, and there is a constant C for which

0< limC,=C<

n-m

This theorem clearly implies the following theorem, which is known

as the integral test for convergence of series.

Theorem 12.251 (integral test) If f is positive and continuous and

decreasing over the interval x > 1, if f (x) -* 0 as x --> -, and if Uk = f (k)

for each k = 1, 2, 3, , then Zuk < if and only if f' f(x) dx < -.

It can be shown that C, >= C2? C3 - - ,and this result is some-

times useful. The most important application of Theorem 12.25 involves

the case in which f(x) = 1/x, Euk is the harmonic series, and the constants

C. and C are called y,, and y (gamma). This application gives

(12.26)

n

1 = log n + 'Yn,

where the constant y for which

(12.261) y = lim yn = 0.57721 56649 01532 86061

n- -

is called the Euler constant. This constant y is, after 7r and e, the most

important mathematical constant not appearing in elementary arithmetic.

Putting f(x) = 1/x", where s > 1, gives

(12.271) ,1 ka = s^11 [1 - nl lJ + C,(s),