Understanding Engineering Mathematics

So the series converges, as we already know from Problem 14.12. It is in fact the

series fore^1 −1.

(ii) For 1+

2

1

+

22

2!

+

23

3!

+

24

4!

+

25

5!

+···we haveun=

2 n

n!

.

So

lim

n→∞

∣

∣

∣

∣

un+ 1

un

∣

∣

∣

∣=nlim→∞

∣

∣

∣

∣

2 n+^1

(n+ 1 )!

n!

2 n

∣

∣

∣

∣=nlim→∞^2

∣

∣

∣

∣

1

n+ 1

∣

∣

∣

∣=^0 <^1

Therefore the series converges.

(iii) For the series 1+

1

22

+

1

32

+

1

42

+

1

52

+···we have

lim

n→∞

∣

∣

∣∣un+^1

un

∣

∣

∣∣= lim

n→∞

∣

∣

∣∣^1

(n+ 1 )^2

n^2

∣

∣

∣∣= 1

So in this case the ratio test isinconclusive. However, it is possible to extend the

method used for the harmonic series to prove that the above seriesconvergesand in

fact more generally, the series

1 +

1

2 p

+

1

3 p

+

1

4 p

+

1

5 p

+···

converges ifp>1 and diverges ifp≤1. This series provides a good standard series

for use in a comparison test.

Absolute convergence

The infinite seriesu 1 +u 2 +u 3 +···is said to beabsolutely convergentif the corre-
sponding series of positive terms|u 1 |+|u 2 |+|u 3 |+···is convergent. If the seriesu 1 +
u 2 +u 3 +···is convergent, but the series|u 1 |+|u 2 |+|u 3 |+···is divergent, we say the
series isconditionally convergent. For example:

1 −

1

22

+

1

32

−···is absolutely convergent (see Problem 14.14(iii)).

1 −^12 +^13 −^14 +···is conditionally convergent (by the harmonic series).

The real power of absolute convergence comes in when we consider infinite power series.
If a power series inxis absolutely convergent, then we can differentiate or integrate it
with respect tox– term by term, which may not be possible otherwise.

Convergence testing summary

Below we summarise a systematic approach to testing the convergence of a series – not
foolproof, but a good start.
When testing the series:

S=u 1 +u 2 +u 3 +u 4 +···

follow the steps below:

(i) Check thatun→0asn→∞.Ifundoes not tend to zero then the

series diverges.