Convergence of infinite series

SERIES AND LIMITS
Again using the Maclaurin expansion of exp x given in subsection 4.6.3, we notice that
S(θ) = Re [exp(exp iθ)] = Re [exp(cos θ + i sin θ)]
= Re {[exp(cos θ)][exp(i sin θ)]} = [exp(cos θ)]Re [exp(i sin θ)]
= [exp(cos θ)][cos(sin θ)]. 4.3 Convergence of infinite series
Although the sums of some commonly occurring infinite series may be found,
the sum of a general infinite series is usually difficult to calculate. Nevertheless,
it is often useful to know whether the partial sum of such a series converges to
a limit, even if the limit cannot be found explicitly. As mentioned at the end of
section 4.1, if we allow N to tend to infinity, the partial sum
SN =
N
un
n=1
of a series may tend to a definite limit (i.e. the sum S of the series), or increase
or decrease without limit, or oscillate finitely or infinitely.
To investigate the convergence of any given series, it is useful to have available
a number of tests and theorems of general applicability. We discuss them below;
some we will merely state, since once they have been stated they become almost
self-evident, but are no less useful for that.
4.3.1 Absolute and conditional convergence
Let us first consider some general points concerning the convergence, or otherwise,
of an infinite series. In general an infinite series
un can have complex terms,
and in the special case of a real series the terms can be positive or negative. From
any such series, however, we can always construct another series
|un | in which
each term is simply the modulus of the corresponding term in the original series.
Then each term in the new series will be a positive real number.
un also converges, and
un is said to be
If the series
|un | converges then
absolutely convergent, i.e. the series formed by the absolute values is convergent.
For an absolutely convergent series, the terms may be reordered without affecting
un converges
the convergence of the series. However, if
|un | diverges whilst
then
un is said to be conditionally convergent. For a conditionally convergent
series, rearranging the order of the terms can affect the behaviour of the sum
and, hence, whether the series converges or diverges. In fact, a theorem due
to Riemann shows that, by a suitable rearrangement, a conditionally convergent
series may be made to converge to any arbitrary limit, or to diverge, or to oscillate
finitely or infinitely! Of course, if the original series
un consists only of positive
real terms and converges then automatically it is absolutely convergent.
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4.3 CONVERGENCE OF INFINITE SERIES
4.3.2 Convergence of a series containing only real positive terms
As discussed above, in order to test for the absolute convergence of a series
un , we first construct the corresponding series
|un | that consists only of real
positive terms. Therefore in this subsection we will restrict our attention to series
of this type.
We discuss below some tests that may be used to investigate the convergence of
such a series. Before doing so, however, we note the following crucial consideration.
In all the tests for, or discussions of, the convergence of a series, it is not what
happens in the first ten, or the first thousand, or the first million terms (or any
other finite number of terms) that matters, but what happens ultimately.
Preliminary test
A necessary but not sufficient condition for a series of real positive terms
un
to be convergent is that the term un tends to zero as n tends to infinity, i.e. we
require
lim un = 0.
n→∞
If this condition is not satisfied then the series must diverge. Even if it is satisfied,
however, the series may still diverge, and further testing is required.
Comparison test
The comparison test is the most basic test for convergence. Let us consider two
vn and suppose that we know the latter to be convergent (by
series
un and
some earlier analysis, for example). Then, if each term un in the first series is less
than or equal to the corresponding term vn in the second series, for all n greater
than some fixed number N that will vary from series to series, then the original
vn is convergent and
series
un is also convergent. In other words, if
un ≤ vn
for n > N,
then
un converges.
However, if
vn diverges and un ≥ vn for all n greater than some fixed number
then
un diverges.
Determine whether the following series converges:
∞
n=1
1
1 1 1
1
= + + +
+ ··· .
n! + 1
2 3 7 25
(4.7)
Let us compare this series with the series
∞
1
1
1
1
1
1
1
=
+
+
+
+ ··· = 2 +
+
+ ··· ,
n!
0! 1! 2! 3!
2! 3!
n=0
125
(4.8)
SERIES AND LIMITS
which is merely the series obtained by setting x = 1 in the Maclaurin expansion of exp x
(see subsection 4.6.3), i.e.
1
1
1
+
+
+ ··· .
1! 2! 3!
Clearly this second series is convergent, since it consists of only positive terms and has a
finite sum. Thus, since each term un in the series (4.7) is less than the corresponding term
1/n! in (4.8), we conclude from the comparison test that (4.7) is also convergent. exp(1) = e = 1 +
D’Alembert’s ratio test
The ratio test determines whether a series converges by comparing the relative
magnitude of successive terms. If we consider a series
un and set
un+1
ρ = lim
,
(4.9)
n→∞
un
then if ρ < 1 the series is convergent; if ρ > 1 the series is divergent; if ρ = 1
then the behaviour of the series is undetermined by this test.
To prove this we observe that if the limit (4.9) is less than unity, i.e. ρ < 1 then
we can find a value r in the range ρ < r < 1 and a value N such that
un+1
< r,
un
for all n > N. Now the terms un of the series that follow uN are
uN+1 ,
uN+2 ,
uN+3 ,
...,
and each of these is less than the corresponding term of
ruN ,
r 2 uN ,
r 3 uN ,
... .
(4.10)
However, the terms of (4.10) are those of a geometric series with a common
ratio r that is less than unity. This geometric series consequently converges and
therefore, by the comparison test discussed above, so must the original series
un . An analogous argument may be used to prove the divergent case when
ρ > 1.
Determine whether the following series converges:
∞
1
1
1
1
1
1
1
=
+
+
+
+ ··· = 2 +
+
+··· .
n!
0! 1! 2! 3!
2! 3!
n=0
As mentioned in the previous example, this series may be obtained by setting x = 1 in the
Maclaurin expansion of exp x, and hence we know already that it converges and has the
sum exp(1) = e. Nevertheless, we may use the ratio test to confirm that it converges.
Using (4.9), we have
n!
1
ρ = lim
= lim
=0
(4.11)
n→∞ (n + 1)!
n→∞
n+1
and since ρ < 1, the series converges, as expected. 126
4.3 CONVERGENCE OF INFINITE SERIES
Ratio comparison test
As its name suggests, the ratio comparison test is a combination of the ratio and
comparison tests. Let us consider the two series
un and
vn and assume that
we know the latter to be convergent. It may be shown that if
vn+1
un+1
≤
un
vn
for all n greater than some fixed value N then
Similarly, if
un is also convergent.
un+1
vn+1
≥
un
vn
for all sufficiently large n, and
vn diverges then
un also diverges.
Determine whether the following series converges:
∞
n=1
1
1
1
= 1 + 2 + 2 + ··· .
(n!)2
2
6
In this case the ratio of successive terms, as n tends to infinity, is given by
R = lim
n→∞
n!
(n + 1)!
2
= lim
n→∞
1
n+1
2
,
which is less than the ratio seen in (4.11). Hence, by the ratio comparison test, the series
converges. (It is clear that this series could also be found to be convergent using the ratio
test.) Quotient test
The quotient test may also be considered as a combination of the ratio and
comparison tests. Let us again consider the two series
un and
vn , and define
ρ as the limit
ρ = lim
n→∞
un
vn
.
(4.12)
Then, it can be shown that:
vn either both converge or both
(i) if ρ = 0 but is finite then
un and
diverge;
un converges;
(ii) if ρ = 0 and
vn converges then
(iii) if ρ = ∞ and
vn diverges then
un diverges.
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SERIES AND LIMITS
Given that the series
∞
n=1
1/n diverges, determine whether the following series converges:
∞
4n2 − n − 3
.
n3 + 2n
n=1
(4.13)
If we set un = (4n2 − n − 3)/(n3 + 2n) and vn = 1/n then the limit (4.12) becomes
2
3
(4n − n − 3)/(n3 + 2n)
4n − n2 − 3n
= lim
= 4.
ρ = lim
3
n→∞
n→∞
1/n
n + 2n
Since ρ is finite but non-zero and
vn diverges, from (i) above
un must also diverge. Integral test
The integral test is an extremely powerful means of investigating the convergence
of a series
un . Suppose that there exists a function f(x) which monotonically
decreases for x greater than some fixed value x0 and for which f(n) = un , i.e. the
value of the function at integer values of x is equal to the corresponding term
in the series under investigation. Then it can be shown that, if the limit of the
integral
N
f(x) dx
lim
N→∞
exists, the series
un is convergent. Otherwise the series diverges. Note that the
integral defined here has no lower limit; the test is sometimes stated with a lower
limit, equal to unity, for the integral, but this can lead to unnecessary difficulties.
Determine whether the following series converges:
∞
n=1
1
4
4
=4+4+ +
+ ··· .
(n − 3/2)2
9 25
Let us consider the function f(x) = (x − 3/2)−2 . Clearly f(n) = un and f(x) monotonically
decreases for x > 3/2. Applying the integral test, we consider
N
1
−1
= 0.
dx = lim
lim
2
N→∞
N→∞
(x − 3/2)
N − 3/2
Since the limit exists the series converges. Note, however, that if we had included a lower
limit, equal to unity, in the integral then we would have run into problems, since the
integrand diverges at x = 3/2. The integral test is also useful for examining the convergence of the Riemann
zeta series. This is a special series that occurs regularly and is of the form
∞
1
.
np
n=1
It converges for p > 1 and diverges if p ≤ 1. These convergence criteria may be
derived as follows.
128
4.3 CONVERGENCE OF INFINITE SERIES
Using the integral test, we consider
1−p N
N
1
dx
=
lim
lim
,
N→∞
N→∞ 1 − p
xp
and it is obvious that the limit tends to zero for p > 1 and to ∞ for p ≤ 1.
Cauchy’s root test
Cauchy’s root test may be useful in testing for convergence, especially if the nth
terms of the series contains an nth power. If we define the limit
ρ = lim (un )1/n ,
n→∞
then it may be proved that the series
un converges if ρ < 1. If ρ > 1 then the
series diverges. Its behaviour is undetermined if ρ = 1.
Determine whether the following series converges:
∞ n
1
1
1
+ ··· .
=1+ +
n
4
27
n=1
Using Cauchy’s root test, we find
1
= 0,
n→∞
n
ρ = lim
and hence the series converges. Grouping terms
We now consider the Riemann zeta series, mentioned above, with an alternative
proof of its convergence that uses the method of grouping terms. In general there
are better ways of determining convergence, but the grouping method may be
used if it is not immediately obvious how to approach a problem by a better
method.
First consider the case where p > 1, and group the terms in the series as
follows:
1
1
1
1
1
+ p +
+ ··· + p + ··· .
SN = p +
1
2p
3
4p
7
Now we can see that each bracket of this series is less than each term of the
geometric series
1
2
4
SN = p + p + p + · · · .
1
2
4
p−1
; since p > 1, it follows that
This geometric series has common ratio r = 12
r < 1 and that the geometric series converges. Then the comparison test shows
that the Riemann zeta series also converges for p > 1.
129
SERIES AND LIMITS
The divergence of the Riemann zeta series for p ≤ 1 can be seen by first
considering the case p = 1. The series is
1 1 1
+ + + ··· ,
2 3 4
which does not converge, as may be seen by bracketing the terms of the series in
groups in the following way:
N
1
1 1
1 1 1 1
+
+ + +
un = 1 +
SN =
+
+
+ ··· .
2
3 4
5 6 7 8
SN = 1 +
n=1
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 that SN increases indefinitely as N is increased.
Now returning to the case of the Riemann zeta series for p < 1, we note that
each term in the series is greater than the corresponding one in the series for
which p = 1. In other words 1/np > 1/n for n > 1, p < 1. The comparison test
then shows us that the Riemann zeta series will diverge for all p ≤ 1.
4.3.3 Alternating series test
The tests discussed in the last subsection have been concerned with determining
un
whether the series of real positive terms
|un | converges, and so whether
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. an alternating series.
An alternating series can be written as
∞
(−1)n+1 un = u1 − u2 + u3 − u4 + u5 − · · · ,
n=1
with all un ≥ 0. Such a series can be shown to converge provided (i) un → 0 as
n → ∞ and (ii) un < un−1 for all n > N for some finite N. If these conditions are
not met then the series oscillates.
To prove this, suppose for definiteness that N is odd and consider the series
starting at uN . The sum of its first 2m terms is
S2m = (uN − uN+1 ) + (uN+2 − uN+3 ) + · · · + (uN+2m−2 − uN+2m−1 ).
By condition (ii) above, all the parentheses are positive, and so S2m increases as
m increases. We can also write, however,
S2m = uN − (uN+1 − uN+2 ) − · · · − (uN+2m−3 − uN+2m−2 ) − uN+2m−1 ,
and since each parenthesis is positive, we must have S2m < uN . Thus, since S2m
130