Mathematical Methods for Physics and Engineering : A Comprehensive Guide

SERIES AND LIMITS

The divergence of the Riemann zeta series forp≤1 can be seen by first

considering the casep= 1. The series is

SN=1+

1

2

+

1

3

+

1

4

+···,

which doesnotconverge, as may be seen by bracketing the terms of the series in

groups in the following way:

SN=

∑N

n=1

un=1+

(

1

2

)

+

(

1

3

+

1

4

)

+

(

1

5

+

1

6

+

1

7

+

1

8

)

+···.

The sum of the terms in each bracket is≥^12 and, since as many such groupings

can be made as we wish, it is clear thatSNincreases indefinitely asNis increased.

Now returning to the case of the Riemann zeta series forp<1, we note that

each term in the series is greater than the corresponding one in the series for

whichp=1.Inotherwords1/np> 1 /nforn>1,p<1. The comparison test

then shows us that the Riemann zeta series will diverge for allp≤1.

4.3.3 Alternating series test

The tests discussed in the last subsection have been concerned with determining

whether the series of real positive terms

∑

|un|converges, and so whether

∑

un

is absolutely convergent. Nevertheless, it is sometimes useful to consider whether

a series is merely convergent rather than absolutely convergent. This is especially

true for series containing an infinite number of both positive and negative terms.

In particular, we will consider the convergence of series in which the positive and

negative terms alternate, i.e. analternating series.

An alternating series can be written as

∑∞

n=1

(−1)n+1un=u 1 −u 2 +u 3 −u 4 +u 5 −···,

with allun≥0. Such a series can be shown to converge provided (i)un→0as

n→∞and (ii)un<un− 1 for alln>Nfor some finiteN. If these conditions are

not met then the series oscillates.

To prove this, suppose for definiteness thatNis odd and consider the series

starting atuN. The sum of its first 2mterms is

S 2 m=(uN−uN+1)+(uN+2−uN+3)+···+(uN+2m− 2 −uN+2m− 1 ).

By condition (ii) above, all the parentheses are positive, and soS 2 mincreases as

mincreases. We can also write, however,

S 2 m=uN−(uN+1−uN+2)−···−(uN+2m− 3 −uN+2m− 2 )−uN+2m− 1 ,

and since each parenthesis is positive, we must haveS 2 m<uN. Thus, sinceS 2 m