162 CHAPTER 5 ALGEBRA

an expression, with whatever method works (adding zero, multiplying by one, adding

or subtracting a bit, etc.), in order to make it more manageable (yet another idea in­

spired by the wishful thinking strategy). Here is another example.

Example 5.3.5 The zeta function �(s) is defined by the infinite series

I 1 1

�(s): = -+-+-+ ....

IS 2 S^3 s

When s = I, this becomes the harmonic series and diverges.^8

Show that �( s) converges for all s 2 2.

Solution: This is a routine exercise using the integral test from calculus, but it is

much more instructive to use first principles. First of all, we note that even though the

problem asks us to prove a statement about infinitely many values of s, we need only

show that �( 2 ) < 00, since if s > 2 we have

1 1

->­

k 2 - kS

for all positive integers k and consequently the convergence of �( (^2) ) will imply the
convergence of �(s) for all larger values of s.
But how to show that �(2) converges? The general term is l/k^2 ; we must search
for a similar series that we know something about. We already "own" a nicely telescop­
ing series whose terms are the reciprocals of quadratics, Example 5.3. 1 on page 158 ,
the series
n Il
�
! k( k +^1 )
= 1 -
n + 1
.
Certainly the infinite series converges (to 1), and now we would be done if 1/ k^2 were
less than 1/(k(k+ (^1) )). But the inequality goes the wrong way!
Not to worry: We can write shift the summation index and get
Now we can use
n Il n+! 1
�
!k(k+l)

2:

• �
  2
  ( k-l)k'
  1 1
  k 2
  <
  k(k-l)'
  for all positive integers k, to conclude that
  n Il
  k
  �
  2
  k^2
  < 1 -
  n + 1.
  So the sum (starting from k = 2) converges, and therefore the entire sum converges as
  well. _
  SIt turns out that '(s) has many rich properties, with wonderful connections to combinatorics and number
  theory. Consult Chapter 2 of [47] as a starting point.