Series
If we add the terms of an infinite sequanece {an}, we get an expression of the form
a1 + a2 + ... + an
which is called an infinite series (or just series), and is denoted by the formal expression
We consider the partial sums
s0 = a0
s0 = a0 + a1
s0 = a0 + a1 + a2
sn = a0 + a1 + ... + an
and, in general,
These partial sums form a new sequence {sn}. If limn ⟶ ∞ sn = s exists (as a finite number), then we call it the sum of the infinite series Σ an.
Definition 3.1.1 With respect to series 3.1.1 considering the limit limn ⟶ ∞ sn, we can have the following situations:
If the limit exists and is finite, we say that the series converges; The value s of the limit is called sum of the series, and we write:
If the limit is ±∞ we say the series diverges;
If the limit does not exist, we say the series is indeterminate. □
The basic necessary and sufficient condition for convergence, due to Cauchy, is a simple reformulation in terms of series of Cauchy's criterion' for convergence of a sequence, which we introduced for sequences. We simply have to refer to the sequence of partial sums {sn} of an infinite series Σ an, giving the following criterion for series.
Theorem 3.1.2 (Cauchy’s convergence criterion for series). Let Σ an be an infinite series with a sequence of partial sums {sn}. The series converges if and only if its sequence {sn} is a Cauchy sequence.
Proof. Suppose the Cauchy criterion holds. Then for every ε > 0, there is an integer N ∈ ℕ such that
Thus the sequence {sn} is Cauchy in the sense discussed for sequences. We conclude that the sequence {sn} converges; by definition, therefore, the series converges.
Conversely, if the series converges then, by definition 3.1.1, the sequence {sn} of partial sums converges. In particular, the sequence {sn} must be Cauchy. Thus, for any ε > 0, there is a number N > 0 such that
n ≥ m < N ⟶ |sn − sm| < ε
This just says that
and this last inequality is the Cauchy criterion for series. □
From the theorem a simple necessary condition for convergence follows. This condition does not guarantee the convergence of the series, but if not satisfied it excludes it.
Corollary 3.1.3. (The zero test) If the series Σ an is convergent, then limn ⟶ ∞ an = 0
Proof. Since we are assuming that the series converges, it must satisfy Cauchy criterion. Let ε > 0. Then there is an integer N ≥ 1 such that if n ≥ m > N, then 3.1.2 holds; If we take n = m and m > N, 3.1.2 becomes
|am| < ε
which proves the claim. □
Cauchy criterion can be states succinctly as: an infinite series converges if and only if it is Cauchy sequence.
Example 3.1.4. The harmonic series is defined by
Using Cauchy's criterion one deduces that the harnomic series diverges. Indeed, the sequence {sn} if its partial sums verifies
thus the Cauchy criterion fails for any ε < 1/2; the harmonic series diverges. ■
The divergence of the harmonic series was first established by the Italian mathematician Mengoli in 1650.
Example 3.1.3 (Mengoli's series)
Observe that 1/n(n +1) = 1/n − 1/(n+1), we are able to give a simpole expression of the sequnce sn
By noting that equal terms of opposite sign can be coupled and vanish. Thus sn ⟶ 1, hence the series converges and has sum equal to 1. ■
Example 3.1.2. (Geometric progression and series). A sequence is called a geometric progression if each term in the sequence (after the first) is a constant multiple of the previous term.
a, aq, aq2, aq3, ... , aqn
where a is said first term, and q said common ratio
Thus, if the terms are {ai} for i = 0, 1, 2, 3..., then ai+1 = qai for some constant, . For example
Progression | Common ratio |
---|---|
1,3,9,27,81, ... | 3 |
16/27, −8/9, 4/3, −2, ... | −3/2 |
x, x2, x3, x4 ... | x |
The nth term of a geometric sequence whose first term is a and common ratio is a non-zero number q is
an = aqn−1
A geometric series is any series that can be put into the form
with q ∈ ℝ. The partial sums of a geometric series with q = 1, then sn = a + a + ... + a = a(n + 1); as n ⟶ ∞ the series diverges in this case.
If instead q ≠ 1, we have
sn = a + aq + aq2 + ... + aqn + ...
and
qsn = qa + aq2 + aq3 + ... + aqn+1 + ...
Subtracting these equations, we get
sn − qsn = a − aqn+1
Thus
see the exercise on mathematical induction for a rigorous proof of the previous relation.
Considering the limit for n ⟶ ∞ of sn, we have