Mathematical prerequisites
2.4 Sequences and series .1 Basic properties
The concept of an infinite series plays a predominant role in chapters 3 to 5 and will also return regularly later on. We will assume that the reader is already acquainted with the theory of sequences and series, as far as the terms of the sequence or the series consist of real numbers. In this section the theory is extended to complex numbers. In general, the terms of a sequence
(an) with n= 0, 1, 2, . . .
will then be complex numbers, as is the case for the terms of a series ∞
n=0
an= a0+ a1+ · · · .
For limits of sequences of complex numbers we follow the same line as for limits of complex-valued functions. Assuming that the concept of convergence of a sequence of real numbers is known, we start with the following definition of convergence of a sequence of complex numbers.
A sequence(an) of complex numbers with un = Re anand vn= Im anconverges DEFINITION 2.5
Convergence of sequences if both the sequence of real numbers(un) and the sequence of real numbers (vn) converge. Moreover, the limit of the sequence(an) then equals
nlim→∞an= lim
n→∞un+ i lim n→∞vn. Let the sequence(an) be given by EXAMPLE 2.14
an= n(ei/n− 1) with n= 0, 1, 2, . . ..
Since ei/n = cos(1/n) + i sin(1/n) one has un = Re an = n(cos(1/n) − 1) and vn = Im an = n sin(1/n). Verify for yourself that limn→∞un = 0 and limn→∞vn= 1. Hence, the sequence (an) converges and limn→∞an= i. Our next theorem resembles theorem 2.3 and can also be proven in the same way.
A sequence(an) converges and has limit a if and only if limn→∞| an− a | = 0.
THEOREM 2.6
For complex z one has EXAMPLE 2.15
nlim→∞zn= 0 if| z | < 1.
We know that for real r with−1 < r < 1 one has limn→∞rn= 0. So if | z | < 1, then limn→∞zn = limn→∞| z |n = 0. Using theorem 2.6 with a = 0 we
conclude that limn→∞zn= 0.
In the complex plane| an− a | is the distance from anto a. Theorem 2.6 states that for convergent sequences this distance tends to zero as n tends to infinity.
A sequence diverges, or is called divergent, if the sequence does not converge.
Divergence of sequences
All kinds of properties that are valid for convergent sequences of real numbers are valid for convergent sequences of complex numbers as well. We will formulate the following properties, which we immediately recognize from convergent sequences of real numbers:
Let(an) and (bn) be two convergent sequences such that limn→∞an = a and limn→∞bn= b. Then
a limn→∞(αan+ βbn) = αa + βb for all α, β ∈ C, b limn→∞anbn= ab,
c limn→∞an/bn= a/b if b = 0.
Now that we know what a convergent sequence is, we can define convergence of a series in the usual way. For this we use the partial sums snof the sequence(an):
Partial sum
sn= n k=0
ak= a0+ a1+ · · · + an.
A series∞
n=0anis called convergent if and only if the sequence of partial sums DEFINITION 2.6
Convergence of a series (sn) converges.
When s = limn→∞sn, we call s the sum of the series. For a convergent series Sum of a series
the sum is also denoted by∞
n=0an, so s=∞ n=0an. Consider the geometric series∞
n=0znwith ratio z ∈ C. The partial sum sn is EXAMPLE 2.16
Geometric series equal to sn = 1 + z + z2+ · · · + zn. Note that this is a polynomial of degree n. When z = 1, then we see by direct substitution that sn = n + 1. Hence, the geometric series diverges for z= 1, since limn→∞sn= ∞. Multiplying snby the factor 1− z gives
(1 − z)sn= 1 + z + z2+ · · · + zn− z(1 + z + z2+ · · · + zn) = 1 − zn+1. For z= 1 one thus has
sn= 1− zn+1 1− z .
For| z | < 1 we have seen that limn→∞zn = 0 (see example 2.15); so then the series converges with sum equal to 1/(1 − z). We write
∞ n=0
zn= 1
1− z if| z | < 1.
Since for| z | ≥ 1 the sequence with terms zn does not tend to zero, the series
diverges for these values of z.
Just as for sequences, the convergence of a series can be verified on the basis of the real and imaginary parts of the terms. For sequences we used this as a definition.
For series we formulate it as a theorem.
Let(an) be a sequence of numbers and un= Re anand vn= Im an. Then the series
THEOREM 2.7 ∞
n=0anconverges if and only if both the series∞
n=0unand the series∞ n=0vn converge.
Proof
Let(sn), (rn) and (tn) be the partial sums of the series with terms an, un and vn respectively. Note that sn= rn+ itn. According to the definition of convergence of a sequence, the sequence(sn) converges if and only if both the sequence (rn) and the sequence(tn) converge. This is precisely the definition of convergence for the
series with terms an, unand vn.
From the preceding theorem we also conclude that for a convergent series with terms an= un+ ivnone has
∞ n=0
an= ∞ n=0
un+ i ∞ n=0
vn.
For convergent series with complex terms, the same properties hold as for conver-gent series with real terms. Here we formulate the linearity property, which is a direct consequence of definition 2.6 and the linearity property for series with real terms.
Let∞
n=0anand∞
n=0bnbe convergent series with sum s and t respectively. Then THEOREM 2.8
∞ n=0
(αan+ βbn) = αs + βt for allα, β ∈ C.
The next property formulates a necessary condition for a series to converge.
If the series∞
n=0anconverges, then limn→∞an= 0.
THEOREM 2.9
Proof
If the series∞
n=0anconverges and has sum s, then nlim→∞an= lim
n→∞(sn− sn−1) = lim
n→∞sn− lim
n→∞sn−1= s − s = 0.
As a consequence we have that limn→∞an = 0 excludes the convergence of the series. The theorem only gives a necessary and not a sufficient condition for convergence. To show this, we consider the harmonic series.
Series of the type∞
n=11/np, with p a real constant, are called harmonic series.
EXAMPLE 2.17
Harmonic series From the theory of series with real terms it is known that for p> 1 the harmonic series is convergent, while for p≤ 1 the harmonic series is divergent. Hence, for 0< p ≤ 1 we obtain a divergent series with terms that do tend to zero.
2.4.2 Absolute convergence and convergence tests
For practical applications one usually needs a strong form of convergence of a se-ries. Convergence itself is not enough, and usually one requires in addition the convergence of the series of absolute values, or moduli, of the terms.
A series∞
n=0an is called absolutely convergent if the series∞
n=0| an| con-DEFINITION 2.7
Absolute convergence verges.
The series of the absolute values is a series with non-negative real terms. Con-vergence of series with non-negative terms can be verified using the following test, which is known as the comparison test.
When(an) and (bn) are sequences of real numbers with 0 ≤ an ≤ bnfor n = THEOREM 2.10
Comparison test 0, 1, 2, . . ., and the series∞
n=0bnconverges, then the series∞
n=0anconverges as well.
Proof
Let(sn) be the partial sums of the series with terms an. Then sn+1−sn= an+1≥ 0.
So sn+1≥ sn, which means that the sequence(sn) is a non-decreasing sequence. It is known, and this is based on fundamental properties of the real numbers, that such a sequence converges whenever it has a upper bound. This means that there should be a constant c such that sn≤ c for all n = 0, 1, 2, . . .. For the sequence (sn) this is easy to show, since it follows from an≤ bnthat
sn = a0+ a1+ · · · + an≤ b0+ b1+ · · · + bn
≤ b0+ b1+ · · · + bn+ bn+1+ · · · .
The sequence(sn) apparently has as upper bound the sum of the series with terms bn. This sum exists since it is given that this series converges. This proves the
theorem.
Absolutely convergent series with real terms are convergent. One can show this as follows. Write an = bn− cn with bn = | an| and cn = | an| − an. It is given that the series with terms bn converges. The terms cnsatisfy the inequality 0≤ cn ≤ 2 | an|. According to the comparison test, the series with terms cnthen converges as well. Since an= bn− cn, the series with terms anthus converges. For series with complex terms, the statement is a consequence of the next theorem.
Let(an) be a sequence of numbers and let un= Re anand vn= Im an. The series
THEOREM 2.11 ∞
n=0anconverges absolutely if and only if both the series∞
n=0unand∞ n=0vn converge absolutely.
Proof
For the terms un and vn one has the inequalities| un| ≤ | an|, | vn| ≤ | an|. If the series with the non-negative terms| an| converges, then we know from the com-parison test that the series with, respectively, terms unand vnconverge absolutely.
Conversely, if the series with, respectively, terms un and vn converge absolutely, then the series with the non-negative terms| un| + | vn| also converges. It then follows from the inequality| an| =
u2n+ v2n≤ | un| + | vn| and again the com-parison test that the series with terms| an| converges. This means that the series
with terms anconverges absolutely.
An absolutely convergent series is convergent.
THEOREM 2.12
Proof
Above we sketched the proof of this statement for series with real terms. For series with, in general, complex terms, the statement follows from the preceding theorem.
Specifically, when the series with complex terms is absolutely convergent, then the series consisting of the real and the imaginary parts also converge absolutely. These are series of real numbers and therefore convergent. Next we can apply theorem 2.7, resulting in the convergence of the series with the complex terms. Of course, introducing absolute convergence only makes sense when there are convergent series which are not absolutely convergent. An example of this is the series∞
n=1(−1)n/n. This series converges and has sum − ln 2 (see (2.26) with t= 1), while the series of the absolute values is the divergent harmonic series with p= 1 (see example 2.17).
We now present some convergence tests already known for series with real terms, but which remain valid for series with complex terms.
If| an| ≤ bnfor n= 0, 1, . . . and∞
n=0bnconverges, then∞
n=0anconverges.
THEOREM 2.13
Proof
According to the comparison test (theorem 2.10), the series with terms | an| converges. The series with terms an is thus absolutely convergent and hence
convergent.
Of course, in order to use the preceding theorem, one should first have available a convergent series with non-negative terms. Suitable candidates are the harmonic se-ries with p> 1 and the geometric series with a positive ratio r satisfying 0 < r < 1 (see example 2.16), and all linear combinations of these two as well.
Consider the series∞
and the harmonic series with terms 1/n2converges. A geometric series has the property that the ratio an+1/an of two consecutive terms is a constant. If, more generally, a sequence has terms with the property that nlim→∞
an+1 an
= L
for some L, then one may conclude, as for geometric series, that the series is abso-lutely convergent if L< 1 and divergent if L > 1. In the case L = 1, however, one cannot draw any conclusion. We summarize this, without proof, in the next theorem.
Let(an) be a sequence of terms unequal to zero with limn→∞an+1/an =L for THEOREM 2.14
D’Alembert’s ratio test some L. Then one has:
a if L< 1, then the series with terms anconverges absolutely;
b if L> 1, then the series with terms andiverges.
Consider the series∞
n=1zn/np. Here z is a complex number and p an arbitrary
Hence, the series is absolutely convergent for| z | < 1 and divergent for | z | > 1.
If| z | = 1, then no conclusions can be drawn from the ratio test. For p > 1 and
| z | = 1 we are dealing with a convergent harmonic series and so the given series converges absolutely. For p ≤ 1 we are dealing with a divergent harmonic series and so the series of absolute values diverges. From this we may not conclude that the series itself diverges. Take for example p= 1 and z = −1, then one obtains the
series with terms(−1)n/n, which is convergent.
2.4.3 Series of functions
In the theory of Fourier series in part 2, and of the z-transform in chapter 18, we will encounter series having terms anthat still depend on a variable. The geometric series in example 2.16 can again serve as an example. Other examples are
∞
The first series is an example of a power series. The partial sums of this series are polynomials in z, so functions defined onC. In section 2.5 we will study these more closely. In this section we will confine ourselves to series of the second type, where
the functions are defined onR or on a part of R. We will thus consider series of the type
∞ n=0
fn(t).
Convergence of such a series depends, of course, on the value of t and then the sum will in general depend on t as well. For the values of t for which the series converges, the sum will be denoted by the function f(t). In this case we write
f(t) = ∞ n=0
fn(t)
and call this pointwise convergence. For each value of t one has a different series Pointwise convergence
for which, in principle, one should verify the convergence. It turns out, however, that in many cases it is possible to analyse the convergence for an entire interval.
Let fn(t) = tn. In example 2.16 it was already shown that∞
n=0tnconverges for EXAMPLE 2.20
| t | < 1, with sum 1/(1 − t). This means that the series∞
n=0fn(t) converges on the interval(−1, 1) and that f (t) = 1/(1 − t). Outside this interval, the series
diverges.
One would like to derive properties of f(t) directly from the properties of the functions fn(t), without knowing the function f (t) explicitly as a function of t.
One could wonder, for example, whether a series may be differentiated term-by-term, so whether f(t) =
fn(t) if f (t) =
fn(t). A simple example will show that this is not always permitted.
Let, for example, fn(t) = sin(nt)/n2, then fn(t) = cos(nt)/n. So for each n > 0 EXAMPLE 2.21
the derivative exists for all t∈ R. However, if we now look at∞
n=1 fn(t) at t = 0, then this equals∞
n=11/n, which is a divergent harmonic series, as we have seen in example 2.17. Although all functions fn(t) are differentiable, f (t) is not. One should also be careful with integration. When, for instance, the functions fn(t) are integrable, then one would like to conclude from this that f (t) is also integrable and that
This is not always the case, as our next example will show.
Let un(t) = nte−nt2 for n = 0, 1, 2, . . . and let fn(t) = un(t) − un−1(t) for EXAMPLE 2.22
n= 1, 2, 3, . . . and f0(t) = 0. Then one has for the partial sums sn(t):
sn(t) = f1(t) + · · · + fn(t)
= u1(t) − u0(t) + u2(t) − u1(t) + · · · + un(t) − un−1(t) = un(t).
The sequence of partial sums converges and has limit f(t) = lim
n→∞sn(t) = lim
n→∞nte−nt2 = 0.
On the interval(0, 1) one thus has, on the one hand, 1
while on the other hand
These results are unequal, hence 1 In order to define conditions such that properties like interchanging the order of summation and integration are valid, one could for example introduce the notion of uniform convergence. This is outside the scope of this book. We will therefore always confine ourselves to pointwise convergence, and in the case when one of the properties mentioned above is used, we will always state explicitly whether it is allowed to do so.
EXERCISES
Use the comparison test to prove the convergence of:
2.16
Determine which of the following series converge. Justify each of your answers.
2.17
Show that the following series of functions converges absolutely for all t:
2.18 ∞
n=0 e2i nt 2n4+ 1.