Understanding Engineering Mathematics

Now, provided− 1 <r<1wehavern→0asn→∞,so

lim

n→∞

Sn=

a

1 −r

Thus, the sum to infinity of the geometric series is

S=

a

1 −r

provide|r|<1. On the other hand, if|r|≥1thenrn→∞asn→∞and so the series
diverges.

Problem 14.11

Show that the infinite arithmetic series

S=aY.aYd/Y.aY 2 d/···

diverges for any value of the common differenced(104➤).

Again, we found in Section 3.2.9 that

Sn=a+(a+d)+(a+ 2 d)+···+(a+(n− 1 )d)

=^12 n( 2 a+(n− 1 )d)

Now, asn→∞,Sn→∞, whatever the value ofd,so

lim

n→∞

Sn=∞

and therefore the arithmetic series diverges.
It is not always possible to use thenth partial sum in this way. For example thenth
partial sum of the harmonic series

S= 1 +

1

2

+

1

3

+

1

4

+···+

1

n

+···

is

Sn= 1 +

1

2

+

1

3

+···+

1

n

In this case there is no simple formula forSn, and so we have to find other ways to test
convergence. We did actually do this in Section 14.6 where it was shown that the harmonic
series diverges. We will look at other tests of convergence in the next section.

Exercise on 14.9

FindSnfor the series

1 +^13 +

( 1

3

) 2

+···+

( 1

3

)r

+···

and hence evaluate the sum to infinity.