Understanding Engineering Mathematics

Answer

3

2

[

1 −

( 1

3

)n]

,^32

14.10 Tests for convergence

As noted in the previous section, if we can calculate the general form forSnthen we can
find limn→∞Snand check convergence directly, but this is not always possible. Fortunately,
a number of different tests have been developed to determine whether a series converges
by considering its general form.
Before we even start to discuss the convergence of a series, we must first check that
limn→∞un=0, because if this is not so then the series cannot converge. To see this
suppose that the series converges. Then from:

un=Sn−Sn− 1

we have
lim
n→∞
(Sn−Sn− 1 )= 0 =lim
n→∞
un

So, if the series converges, limn→∞un=0. However, the converse is not true,
i.e. limn→∞un=0, does not imply that the series converges. Thus, we saw in
Section 14.6 that the harmonic series actually diverges and yet it clearly satisfies

limn→∞un=limn→∞

1

n

= 0. On the other hand, for the arithmetic series (Problem 14.11)

limn→∞un=∞and we do know that this series always diverges.
Having verified that limn→∞un=0, we can continue to apply the tests for convergence
given below. First, a couple of general points:

(1) In studying the convergence of a series it is only the ‘infinite tail’

which is important, not an initial finite number of terms.

(2) Some of the tests given below apply only to series whose terms all

eventually become positive.

The comparison test

This is essentially a generalisation of the method we used for the harmonic series. Suppose
we wish to study the convergence of the series of positive terms:

U=u 1 +u 2 +···+un+···

Then in the comparison test wecomparethis with a known series. Thus, let

V=v 1 +v 2 +···+vn+···

be a known standard series. Then ifur≤cvr,c=a constant and V is convergent,
then so isU.Ifur≥cvr andV diverges, then so doesU. This test is intuitively
obvious.