1. Sequences and limits
Introduction
These notes presuppose some previous experience of sequences and limits, such as
the treatment in MATH115. To define a sequence (an ) = (a1 , a2 , . . .) (of real or complex
numbers) one must somehow specify an for each n. This will usually be done by a formula,
such as an = n + 2 or an = 1/n, but there are other examples, such as: an is the number of
divisors of n.
Another way to specify a sequence is to define an+1 in terms of an , as in functional
iteration (e.g. an+1 = 2/(an + 2), with a0 = 1).
Recall that a sequence (an ) is said to be increasing if a1 ≤ a2 ≤ a3 ≤ . . ., that is,
an ≤ an+1 for all n. It is strictly increasing if an < an+1 for all n. The terms decreasing and
strictly decreasing are defined similarly, and monotonic means either increasing or decreasing.
Examples: ( n1 ) is decreasing, (1 − n1 ) is increasing, and (−1)n−1 is neither.
The sequence (an ) is said to be bounded above if there exists a number M such that
an ≤ M for all n. The number M is then said to be an upper bound of (an ). (Note that it is
not unique: for example, M + 1 is also an upper bound.)
Similarly, (an ) is bounded below if there exists K such that an ≥ K for all n, and K
is then a lower bound of (an ). We say that (an ) is bounded if it is bounded both above and
below.
Any increasing sequence is obviously bounded below, by its first term. The sequence
(2n ) is not bounded above. Examples of bounded sequences are: an = 1/n and an = (−1)n .
An elementary fact (MATH115, result 1.1) is: A sequence (an ) is bounded if and only if
(|an |) is bounded above.
Limits
Recall examples like the following:
Example 1. Find the limits (as n → ∞) of:
n2 + 1
=
2n2 + n + 3
n2 − 1
−n=
n+4
1
In these answers, we assumed that 1/n and 1/n2 tend to 0 as n → ∞. We also assumed
the following general rules: if an → a and bn → b as n → ∞, then:
an + bn → a + b,
an bn → ab,
1
1
→
an
a
if a 6= 0.
We now set ourselves the task of proving these statements beyond doubt. The first step is
to be absolutely clear about the meaning of a limit.
Informally, the statement “an tends to the limit a as n → ∞” means: an is as close as
we like to a once n is large enough. A more exact formulation (the one used in MATH115)
is:
Given any positive number ε (however small), we have |an − a| ≤ ε
once n is large enough.
However, it is desirable to be more precise about the phrase “once n is large enough”.
It means: there exists a number n0 such that the required statement holds for all n ≥ n0
(it’s not important for n0 to be an integer).
Putting all this together, we obtain the final, precise version of the definition of the
statement “an → a as n → ∞”:
given any ε > 0 (however small), there exists a number n0
such that |an − a| ≤ ε for all n ≥ n0 .
This definition, and variations of it to come, is absolutely fundamental to Analysis, the
area of Mathematics dealing with things being close to each other (i.e. limits, continuity,
approximations, etc.). Before proceeding further with this course, it is essential to learn it,
and to understand the reasons leading up to it. It might help to write it out again each time
you come back to the subject.
If a sequence tends to a limit, it is said to be convergent. Another allowable way to
write the statement is “limn→∞ an = a” (though this makes better sense when we have
established that a sequence can only have one limit).
A number of minor comments are helpful in coming to terms with this definition.
(1) The words “however small” are logically redundant; they are just included to remind
you of the underlying point.
(2) “Given any ε > 0” is the same as “for all ε > 0. The overall form is an “all”
statement, just like “for all real x, we have x2 ≥ 0”.
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(3) The statement |an − a| ≤ ε is equivalent to: a − ε ≤ an ≤ a + ε.
(4) Any statement of the form |an − a| ≤ ε must be qualified by saying for which n it
holds (in this case, “for all n ≥ n0 ”).
(5) Arrows come in pairs: an → 0 as n → ∞.
(6) Both the following statements are equivalent to an → a as n → ∞: (i) an − a → 0,
(ii) |an − a| → 0 as n → ∞.
(7) The overall meaning is not changed if we substitute |an − a| < ε for |an − a| ≤ ε.
To see this, apply the original statement to the positive number ε/2: for large enough n, we
have |an − a| ≤ ε/2, hence certainly |an − a| < ε.
(8) n0 depends on ε: a smaller ε (demanding even closer approximation) will require a
larger n0 (meaning n will have to be even larger).
(9) No meaning is attached to a statement of the form “an → bn as n → ∞”. If you
mean “an − bn → 0”, then say so!
(10) If (an ) tends to a as n → ∞, then so does (an−1 ). For: write an−1 = bn . If
|an − a| ≤ ε for all n ≥ n0 , then |bn − a| ≤ ε for all n ≥ n0 + 1.
(11) The negation. “an does not tend to a as n → ∞” means:
Having said that the exact meaning of “limit” is as above, we must stick to our principles: all statements that something tends to a limit must be interpreted as a case of the
definition. A proof of such a statement must (in principle) start with “Choose ε > 0”.
Examples and two elementary results
Example 2. If an = a for all large enough n, then an → a as n → ∞. For any given ε,
we actually have |an − a| = 0 (so certainly < ε) for all large enough n.
Example 3. 1/n → 0 as n → ∞. Because: choose ε > 0.
√
Example 4. 1/ n → 0 as n → ∞. Because:
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Example 5. Let an = (−1)n . Then (an ) does not tend to any limit. Whatever a we try,
the interval [a − 12 , a + 12 ] can’t contain both 1 and −1, so the definition fails for ε = 21 .
Recall that |ab| = |a|.|b| and |a + b| ≤ |a| + |b| (hence e.g. |a| ≤ |b| + |a − b|).
The next two results are the source of many further examples. We shall leave the
constantly recurring phrase “as n → ∞ understood (while remembering that arrows really
come in pairs!).
1.1. If an → a, then can → ca.
Proof. Assume c 6= 0 (or the statement is trivial).
1.2. (The sandwich rule) Suppose that bn ≤ an ≤ cn for all n (or at least for all
sufficiently large n), and that bn → a and cn → a. Then an → a.
Proof.
This is typically used in the following form: if bn → 0 and |an | ≤ bn for all n, then
√
an → 0. We already know that 1/n and 1/ n tend to 0, so we can use either of these as bn .
Example 6. 1/(n2 + 1) → 0, since it is less than 1/n.
Example 7. Let an = 1/n if n is even, 0 if n is odd. Then an → 0 as n → ∞,
because 0 ≤ an ≤ 1/n for all n. This shows that an → a does not mean quite the same as
“an gets closer and closer to a”. It is not required that each term is closer than the previous
one.
Example 8. Let an = (1 + n1 )1/2 .
Example 9. Let an = (n + 1)1/2 − n1/2 .
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(But remember that we’re not allowed to say “(n + 1)1/2 → n1/2 ” !)
Further elementary results
We continue to leave the phrase “as n → ∞” understood.
1.3. Suppose that an → 0 and (bn ) is bounded. Then an bn → 0.
Proof. There exists M > 0 such that |bn | ≤ M for all n.
1.4. If an → a and bn → b, then an + bn → a + b.
Proof. Note that |(an + bn ) − (a + b)| =
1.5. Every convergent sequence is bounded.
Proof. Suppose that an → a as n → ∞. Then there exists n0 such that for n > n0 , we
have a − 1 ≤ an ≤ a + 1.
Of course, the example (−1)n shows that not every bounded sequence is convergent.
1.6. If an → a and bn → b, then an bn → ab.
Proof.
1.7. Suppose that an → a, bn → b and an ≤ bn for all sufficiently large n. Then a ≤ b.
Proof. Suppose that a > b. Define ε by: a = b + 3ε.
In particular, if an → a and an ≤ M for all n, then a ≤ M .
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1.8. If a sequence tends to a limit, then the limit is unique.
Proof.
Recall that if 0 < a < a0 and b > b0 > 0, then
a
a0
< 0.
b
b
1.9. If an → a and a 6= 0, then an 6= 0 for all sufficiently large n, and 1/an → 1/a.
Proof. Let ε0 > 0 (to be chosen later). There exists n0 such that if n ≥ n0 , then
|an − a| ≤ ε0 , hence
|an | ≥ |a| − |a − an | ≥ |a| − ε0 .
By 1.6 and 1.9, if an → a and bn → b 6= 0, then
a
an
→
bn
b
as n → ∞.
Tending to infinity
The statement an → ∞ as n → ∞ means (informally): given any M > 0, we have
an ≥ M once n is large enough. Having (we hope) got used to the definition of a limit, you
should have no difficulty with the precise form:
given any M > 0 (however large), there exists n0 such that an ≥ M for all n ≥ n0 .
Also, we say that an → −∞ as n → ∞ if −an → ∞. Clearly, this amounts to replacing
an ≥ M by an ≤ −M in the above.
Note that ∞ is not a real number. It is used in phrases like “tending to ∞”, not in
isolation.
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1.10. If bn → ∞ as n → ∞ and an ≥ bn for all sufficiently large n, then an → ∞ as
n → ∞.
Proof. Assume an ≥ bn for all n ≥ n1 .
Example 10. Let c > 0 and d be any real number. Then cn − d → ∞ as n → ∞.
This example, together with 1.10, is often what we need to show that sequences tend
to infinity.
Example 11. n2 − 4n → ∞
Example 12. n1/2 → ∞ as n → ∞.
A sequence can tend to infinity without being increasing, as the next example shows.
Example 13. Let
an =
2n if n is even,
n if n is odd
Then an ≥ n for all n, so an → ∞ as n → ∞, but clearly (an ) is not increasing.
Example 14. Let an = (n3 − 2)/(2n2 + 4n + 5). Because of the n3 at the top and the
n2 at the bottom, we would expect an to tend to infinity. We can justify this as follows.
We give two results involving sequences tending to infinity.
1.11. If an → ∞, then
1
→ 0.
an
Proof.
1.12. If an → ∞ and (bn ) is bounded above, then an − bn → ∞.
Proof. There exists K such that bn ≤ K for all n.
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Expressions like ∞ − ∞ and 0.∞ are meaningless! The next example shows that one
can have an → ∞, bn → ∞ and an − bn doing anything, so that there is nothing resembling
a rule that would make any sense of “∞ − ∞”
Example 15.
Sequences of powers; applications of the binomial theorem
We shall consider the behaviour of sequences like (an ) for a fixed a. Of course, if a = 1,
this is just the constant sequence 1. The interesting cases are a > 1 and |a| < 1. We shall
use the binomial theorem:
n(n − 1) 2
n r
(1 + h) = 1 + nh +
h + ··· +
h + · · · + hn .
r
2
n
1.13 PROPOSITION. (i) If a > 1, then an tends to ∞ as n → ∞. Moreover, so does
an /nk for any positive integer k.
(ii) If |b| < 1, then bn tends to 0 as n → ∞. Moreover, so does nk bn for any k.
Proof. (i) Write a = 1 + h, so that h > 0. Then
an = (1 + h)n ≥
so an → ∞ as n → ∞, by 1.10. Now let k be given. For n > k, we have in the same way
an ≥
(ii) This is trivial if b = 0, and the case b < 0 follows from the case b > 0, since
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|bn | = |b|n . So assume 0 < b < 1.
Note. This statement can be rephrased informally as saying that “geometric growth
beats polynomial growth”, but such words are indeed only a rephrasing of the result once it
is proved, not in themselves a proof!
Example 16.
1.01n
n3n
→ ∞ and
= n(3/4)n → 0 as n → ∞.
100n
4n
A second result applies to nth roots (assuming their existence, to be proved later).
1.14. For any k > 0, k 1/n → 1 as n → ∞. Also, n1/n → 1 as n → ∞.
Proof.
Complex numbers
If the numbers an and a0 are complex, then |an − a0 | still exists, and still describes the
distance from an to a0 . Hence the meaning of the statement “an → a0 as n → ∞” is defined
exactly as before. Results 1.1 to 1.9 still apply, with no change to the proof, except 1.2 and
1.7, which involve statements like a ≤ b, which have no meaning for complex numbers. Also,
we have the following simple relationship (beware the change in notation!):
1.15. Let zn = an + ibn (where an , bn are real) for n = 0, 1, 2, . . .. Then the following
statements are equivalent:
(i)
(ii)
zn → z0 as n → ∞,
an → a0 and bn → b0 as n → ∞.
Proof.
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Monotonic sequences
We assume the completeness property of the real number system. It says:
If a set E of real numbers is bounded above, then it has a least upper bound (also
called its supremum, denoted by sup(E) ). Similarly, if E is bounded below, then it has a
greatest lower bound (called its infimum, inf(E) ).
The second statement can be deduced from the first, by considering {−x : x ∈ E}.
For M to be the supremum of E means:
(i) M is an upper bound of E (so that x ≤ M for all x in E),
(ii)
if r < M , then r is not an upper bound of E, so there exists x in E with x > r.
Clearly, the supremum of (0, 1) is 1 and the infimum is 0. The same is true for the
closed interval [0, 1]. Note that the supremum of E need not belong to E, but if it does, it
is simply the greatest member of E.
1.16 PROPOSITION. Suppose that (an ) is an increasing sequence. Then:
(i)
(ii)
if (an ) is bounded above, then it converges to its supremum;
if (an ) is not bounded above, then it tends to infinity.
Similar statements apply to decreasing sequences.
Proof.
Note that in case (i), the limit M is the supremum of (an ), hence an ≤ M for all n.
This result is fundamental in the study of series (see section 2).
Functions of a real variable; limits as x → ∞
Let f be a real-valued function of a real variable. Note that we use the notation f for
the function itself, and f (x) for its value at x (hence if f (x) = x2 − 1, then f (3) = 8, etc).
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The limit of a function f (x) as x → ∞ is just like the limit of a sequence (an ) as
n → ∞, with the difference that when we say something like “for all x ≥ R”, we now mean
all real numbers x ≥ R instead of just integers. So the statement “f (x) → ` as x → ∞”
means:
given any ε > 0, there exists R such that |f (x) − `| ≤ ε for all (real) x ≥ R.
(We have replaced an by f (x), a by ` and n0 by R.)
and “f (x) → ∞ as x → ∞” means:
given any M > 0, there exists R such that f (x) ≥ M for all x ≥ R.
Example 17. 1/x → 0 as x → ∞: given ε > 0, we have 0 < 1/x ≤ ε for all x ≥ 1/ε.
(Compare “1/n → 0 as n → ∞”, which was proved in exactly the same way. Only the
convention that x represents a real variable tells us that we now mean for all real x ≥ 1/ε.)
Example 18.
sin x
→ 0 as x → ∞, since
x
A new feature is that x can also tend to −∞. To obtain the definition of “f (x) → `
as x → −∞, we simply replace “for all x ≥ R by “for all x ≤ −R”. For example, 1/x → 0
as x → −∞.
The earlier results for sequences have obvious corresponding versions for functions: for
example, the limits of f (x)+g(x) and f (x)g(x) are what you would expect. Simply replacing
n by x in Examples 1 and 14, we obtain:
Example 19.
x2 − 1
−x
x+4
Example 20.
x3 − 2
→ ∞ as x → ∞.
2x2 + 4x + 5
Recall that a function f is increasing on an interval I if f (x1 ) ≤ f (x2 ) whenever
x1 , x2 ∈ I and x1 < x2 , and strictly increasing on I if f (x1 ) < f (x2 ) for such x1 , x2 .
Example 21. x − 2 sin x → ∞ as x → ∞, since
(Note, however, that this function is not increasing on any
interval [R, ∞).)
We spell out the result analogous to 1.16 (the proof is
still the same):
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1.17. Suppose that f (x) is increasing on [a, ∞) for some a. Then:
(i)
if {f (x) : x ≥ a} is bounded above (with supremum M), then f (x) → M as x → ∞;
(ii)
otherwise, f (x) → ∞ as x → ∞.
1.18 PROPOSITION (behaviour of polynomials). Let p(x) = a0 + a1 x + · · · + an xn ,
where n ≥ 1. Suppose that an > 0. Then:
(ii)
p(x)
→ an as x → ∞ and as x → −∞;
xn
p(x) → ∞ as x → ∞;
(iii)
when x → −∞, p(x) tends to ∞ if n is even, −∞ if n is odd.
(i)
Proof.
A rational function is one of the form f (x) = p(x)/q(x), where p and q are polynomials.
Limits of rational functions are illustrated by Examples 1 and 20.
Subsequences
A subsequence of (an ) is a sequence of the form (an1 , an2 , . . .), i.e. (ank : k = 1, 2, . . .),
where n1 < n2 < n3 < . . .. In other words, we pick out some of the terms of (an ). Obviously,
if an → a as n → ∞, then ank → a as k → ∞.
1.19 PROPOSITION. Every real sequence has a monotonic subsequence.
Proof. This is a short, but clever, proof depending on the following idea. Call n a
“peak point” for the sequence if an > ar for all r > n.
By 1.16 and 1.19, we now have (without further proof) the following important theorem,
which will be used repeatedly later on.
1.20 THEOREM. Every bounded real sequence has a convergent subsequence.
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