Sequences CheatSheet

1 Intro
Sequences CheatSheet
This cheatsheet contains everything you should know about real sequences. It’s not meant to be
exhaustive, but it contains more material than the textbook.
2 Definitions and properties
In this first section we define the notion of sequence of real numbers and the notion of subsequence.
Definition 2.1. A sequence of real numbers an is a function whose domain is the natural numbers
and range is contained in the real numbers:
N a·
R
n
an
Definition 2.2. A sequence an is called increasing if ai < ai+1 for all i ∈ N. It’s called decreasing
if ai > ai+1 for all i ∈ N. It’s called monotonic if it’s either increasing or decreasing.
Definition 2.3. Let p be an increasing sequence of natural numbers (it’s range is contained in N).
Let an be a sequence, we call a subsequence of an the composition apn:
N p· a·
N
R
k
pk
Example 2.4. For example let pk = 2k and an is any sequence; the composition apk = a2k is the
subsequence of all the even members of an. If pk = 2k+1, the composition a2k+1 is the subsequence
of all the odd members.
Definition 2.5. A sequence an is bounded from above if there is a real number L such that an ≤ L
for all n ∈ N. It is bounded from below if there is a real L such that an ≥ L for all n ∈ N. It’s
bounded if it is bounded from above and below.
3 Convergence
A real sequence an is convergent to a real number L if you can make an arbitrarily close to L by
taking n large enough.
apk
Definition 3.1. We say that an converges to L and we denote it with
lim
n→+∞ an = L
if ∀ ɛ ∈ R, ɛ > 0 there exists n(ɛ) ∈ N such that ∀ n > n(ɛ), |an − L| < ɛ.
1

is.
In this definition ɛ measures how close we are to L and n(ɛ) tells us how large “large enough”
Example 3.2. Let an be the sequence 1
np where p is a positive real number. Pick any number ɛ > 0.
The condition:
1
< ɛ
np is fulfilled if:
1
−
n > ɛ p
1
−
The number ɛ p is positive but not natural, so we put
1
−
n(ɛ) = ɛ p
This choice of n(ɛ) proves that an converges to zero. The writing ⌈x⌉ denotes the ceiling function
which maps x to the smallest integer which is greater than or equal to x.
A real sequence an is divergent if you can make an arbitrarily large (positive or negative) by
taking n large enough.
Definition 3.3. We say that an is divergent and denote it with:
lim
n→+∞ an = +∞
if ∀C ∈ R there exists n(C) ∈ N such that ∀ n > n(C), an > C.
Example 3.4. Let an be the sequence ln (n). Let C be any positive real number. The condition
ln (n) > C is satisfied if we pick n > e C because the exponential is an increasing function. We
prove that an is divergent by choosing n(C) = e C .
A sequence might be neither convergent nor divergent; for example an = (−1) n is certainly not
divergent but it doesn’t converge to any number either. In the next section we shall see a method
to prove that a sequence is not convergent.
4 Techniques
Sometimes a sequence can be a real function in disguise. In that case the limit of the sequence is
the limit of the function, if this limit does exist.
Proposition 4.1. Let f(x) be a function of one real variable whose domain contains the interval
[1, +∞). Let an be the sequence f(n) where n is any natural number. If:
lim f(x) = L
x→+∞
where L is a real number or also ±∞, then the sequence an has the same limit:
lim
n→+∞ an = L
If the function f(x) doesn’t have a limit we cannot conclude anything about the sequence: it
might or might not converge. Consider for instance an = sin (πn). The function f(x) = sin (πx)
doesn’t have a limit but an = sin(πn) = 0 for all natural numbers n, so an converges to zero.
There are sequences that do not come from a function of one real variable or they come from
some non elementary function. For instance p n does not come from a real function if p is negative.
The sequence n! comes from a real function, the Gamma function, but unfortunately this function
is not an elementary one. In all these cases we might need some specific technique to study the
convergence.
2

Proposition 4.2. A sequence an converges to L ∈ R if and only if every subsequence converges to
L.
Example 4.3. The sequence an = (−1) n is not convergent. The subsequence a2n converges to
1 and the subsequence a2n+1 converges to −1. If an were convergent to some number L all its
subsequences should converge to the same L, so an is not convergent.
If two subsequences converge to the same limit, in general we cannot conclude that the sequence
is convergent. However the subsequence of even members and the subsequence of odd members are
special: if they are both convergent and they converge to the same limit, then the original sequence
is convergent as well:
Proposition 4.4. Let an be a real sequence. If the subsequence a2n converges to a real number L
and the subsequence a2n+1 converges to the same number L, then an converges to L as well.
Proof. Choose a positive ɛ ∈ R. Since a2n converges to L, there is p(ɛ) such that ∀n > p(ɛ), |a2n −
L| < ɛ. There is also q(ɛ) such that ∀n > q(ɛ), |a2n+1 −L| < ɛ. Set n(ɛ) equal to max(2p(ɛ), 2q(ɛ)+
1). For all n > n(ɛ) we are either in the odd or the even case so |an − L| < ɛ, which proves the
convergence of an to L.
This proposition can be used even when the two subsequences go to infinity.
With the following two propositions we can compare sequences both when they are convergent
and divergent.
Proposition 4.5 (Squeeze theorem). Let an, bn, cn be three sequences of real numbers. If there is
n0 ∈ N such that ∀n ≥ n0, an ≤ bn ≤ cn and both an and cn converge to L ∈ R then also bn
converges to L.
Proof. According to the definition of limit for every ɛ > 0 there exists nɛ such that for every n ≥ nɛ:
and there exists mɛ such that if n ≥ mɛ:
L − ɛ < an < L + ɛ
L − ɛ < cn < L + ɛ
If we now choose p = max nɛ, mɛ, n0 all the inequalities are true simultaneously for every n ≥ p
and we can deduce:
L − ɛ < an ≤ bn ≤ cn < L + ɛ
which proves that bn converges to L.
Proposition 4.6. Let an, bn be two sequences. If there is n0 ∈ N such that ∀n ≥ n0, an ≤ bn and
limn→+∞ an = +∞ then also limn→∞ bn = +∞.
Not as popular as the root ratio test for series, but even sequences have a root and ratio test
that can be helpful in certain calculations:
Proposition 4.7. Let an be a sequence of real numbers such that n |an| converges to L. If L < 1
the sequence converges to zero, if L > 1 the sequence is divergent, if L = 1 the test is inconclusive.
Proposition 4.8. Let an be a sequence of real numbers such that an+1
converges to L. If L < 1
an
the sequence converges to zero, if L > 1 the sequence is divergent, if L = 1 the test is inconclusive.
3

The two tests are perfectly equivalent; if one is inconclusive the other one is also inconclusive.
You can choose the most convenient one in a specific calculation. For example: determine for which
real numbers a the following sequence is convergent:
an = n!an
n n
When dealing with a factorial, the ratio test is always the best bet:
an+1
an
=
(n + 1)a
(n + 1)(1 + 1
n )n
The limit of the sequence is a
e . If a > e the sequence diverges, if a < e the sequence converges to
zero. If a = e the test is inconclusive. It can be shown by finer techniques that the case a = e is
actually divergent.
proof of prop. 4.7. If L < 1 we can choose ɛ > 0 such that L + ɛ < 1. According to the definition
of limit there exists nɛ ∈ N such that if n ≥ nɛ we have:
or equivalently:
L − ɛ < n |an| < L + ɛ < 1
0 ≤ |an| < (L + ɛ) n
Since L + ɛ < 1 the sequence (L + ɛ) n converges to zero and according to the squeeze theorem |an|
converges to zero as well. If the absolute value converges to zero, even an converges to zero.
If L > 1 we can choose ɛ > 0 such that L − ɛ > 1. If n ≥ nɛ we have:
(L − ɛ) n ≤ |an|
and since (L − ɛ) n diverges to +∞, the sequences |an| and an are also divergent.
proof of prop. 4.8. If L < 1 we can choose ɛ > 0 such that L + ɛ < 1. According to the definition
of limit there exists nɛ ∈ N such that if n ≥ nɛ we have:
< L + ɛ < 1
an+1
an
If n ≥ Nɛ we can apply the inequality repeatedly and obtain:
|an| ≤ (L + ɛ) n−nɛ anɛ
0 ≤ |an| < (L + ɛ) n
Since L + ɛ < 1 the sequence (L + ɛ) n−nɛ converges to zero and according to the squeeze theorem
|an| converges to zero as well. If the absolute value converges to zero, even an converges to zero.
If L > 1 we can choose ɛ > 0 such that L − ɛ > 1. If n ≥ nɛ we have:
(L − ɛ) ≤
Applying repeatedly we get:
an+1
an
(L − ɛ) n−nɛ anɛ ≤ |an|
and since (L − ɛ) n−nɛ− diverges to +∞, the sequences |an| and an are also divergent.
4

The following proposition can be helpful when nothing else works.
Proposition 4.9. A real sequence an which is bounded from above and increasing is convergent.
A sequence which is bounded from below and decreasing is convergent.
This statement is a consequence of the least upper bound property of the real numbers. We
introduce here some notation and definitions and we will need to explain this property.
Without entering into formal details, we say that a set X is ordered if the symbol ≤ is defined
on it, and for any two elements x, y ∈ X we can always say whether x ≤ y or y ≤ x.
Definition 4.10. A subset A of an ordered set X is bounded from above if there exists x ∈ X
such that x ≥ a for every a ∈ A. We call x an upper bound of A.
A subset A of an ordered set X is bounded from below if there exists x ∈ X such that x ≤ a
for every a ∈ A. We call x a lower bound of A.
Definition 4.11. The supremum of a bounded from above set A, is the smallest upper bound of
A (it might not exist!). We denote it with sup(A). The infimum of a bounded from below set A,
is the greatest upper bound of A (it might not exist!). We denote it with inf(A).
If α is the supremum of A we can express that it is the smallest upper bound in the following
way:
1. α ≥ a for every a ∈ A
2. For every positive number ɛ, the number α−ɛ is not an upper bound, which means that there
exists aɛ ∈ A such that aɛ > α − ɛ.
Definition 4.12. An ordered set X has the least upper bound property if for every subset A ⊂ X
which is bounded from above, there exists α ∈ X which is the supremum of A.
Theorem 4.13. The real numbers have the least upper bound property.
The proof of this theorem is not obvious and it’s founded on the rigorous definition of the real
numbers that we have accurately omitted.
Remark 4.14. The rational numbers for instance don’t have the least upper bound property. The
set A = {p ∈ Q p 2 < 2} ⊂ Q is clearly bounded from above but it doesn’t have a smallest upper
bound. Such a number would be √ 2 which is not rational.
Remark 4.15. If a set has the least upper bound property then it’s also true that every bounded
from below subset has an infimum.
proof of proposition 4.9. Denote with R the range of the sequence an. Since the sequence is bounded
from above, the set R is also bounded from above. Since the real numbers have the least upper
bound property there is L ∈ R which is the supremum of R. For every ɛ > 0, the number L − ɛ is
not a supremum so there exists nɛ ∈ N such that L − ɛ < anɛ. Since the sequence is increasing, for
every n ≥ nɛ we have:
L − ɛ < anɛ ≤ an < L + ɛ
which proves that L is actually the limit of the sequence.
The proof in the decresing bounded from below case, is exactly the same.
5

5 Induction
Mathematical induction is not a technique specifically meant to study the convergence of a sequence,
nonetheless it can be very helpful in this kind of scenario. We denote with Pn a statement involving
a natural number n. For a given number n the statement can be true or false.
Proposition 5.1. Let P = {Pn; n ∈ N} be a set of statements involving a natural number n.
Suppose that the following conditions are satisfied:
1. the statement P1 is true;
2. Pn is true implies that Pn+1 is true as well;
then the statement Pn is true for every natural number n.
We can replace the first condition with: Pn0 is true; in this case the proposition changes to: the
statement Pn is true for every n ≥ n0.
Example 5.2. We can prove by induction that:
is true for all natural numbers.
Pn :
1. If n = 1 the statement is trivially true.
2. Suppose that Pn is true.
n+1
i =
i=1
n
i + (n + 1) =
i=1
This proves that:
is true.
n
i =
i=1
n(n + 1)
2
Pn+1 :
n(n + 1)
2
+ (n + 1) = n2 + 2n + 3
2
n+1
i =
i=1
(n + 1)(n + 2)
2
According to the proposition, the statement Pn is true for every natural number.
6 Fundamental limits of sequences
= (n + 1)(n + 2)
2
The following limits should be considered fundamental, and you don’t have to retrieve them each
time you do a calculation:
•
lim
n→+∞ np =
⎧
⎪⎨
⎪⎩
6
+∞ if p > 0
1 if p = 0
0 if p < 0

•
•
•
•
•
•
•
lim
n→+∞ pn =
⎧
⎪⎨
⎪⎩
lim
n→+∞
0 if −1 < p < 1
1 if p = 1
+∞ if p > 1
does not exist if p ≤ −1
n√ n = 1
lim 1 +
n→+∞
1
n
lim
n→+∞
n
n!
lim = 0
n→+∞ nn e
lim
n→+∞
n
n!
lim
n→+∞
= 0
np = 0
en = e
ln(n)
= 0 if p > 0
np The number p in all the previous formulas is a real number. The last four limits compare how
fast the functions (n n , n!, e n , n p , ln(n)) go to infinity. The result can be read as: n n is the fastest,
followed by the factorial, followed by the exponential, followed by any power function; and the
logarithm is the slowest.
7 Examples of recursive sequences
A sequence can be defined recursively; this means that we choose the first element in the sequence
and we define any other element as a function of its predecessor:
a1 = c
an+1 = f(an)
where f(·) is a function of one real variable. Sometimes it’s possible to make a recursive sequence
explicit, but it’s usually harder to find the explicit sequence than studying the convergence directly.
Example 7.1. Consider the following recursive sequence:
a1 = 1
an+1 = √ 1 + an
The n-th term of the sequence looks like this:
an = 1 + 1 + √ 1 + . . .
n+1 times
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First of all we notice that this sequence is increasing. We can prove it by induction:
a1 = 1 < a2 = √ 2
Assume that an ≤ an+1 and try to prove that an+1 ≤ an+2:
This implies that:
an = a 2 n+1 − 1 ≤ an+1 = a 2 n+2 − 1
a 2 n+1 ≤ a 2 n+2
and this is satisfied if −an+2 ≤ an+1 ≤ an+2. Now we prove by induction that the sequence is
bounded above by 3. The first element is less then 3. Suppose that an ≤ 3. This implies:
an+1 = √ 1 + an ≤ √ 4 < 3
According to proposition 4.9 the sequence is convergent. Since the sequence is convergent every
subsequence must converge to the same limit, in particular an+1 has the same limit as an:
√
1 + an = √ 1 + L
L = lim
n→+∞ an+1 = lim
n→+∞
This is satisfied if L = √ 1 + L, and since L must be positive L = 1+√5 2
of 1.
Example 7.2. Consider the degree 2 equation
which is the golden ratio
x 2 − bx − c = 0 (1)
where b, c are integers and for simplicity we assume that they are strictly positive. Since x = 0 is
not a solution we can divide everything by x and obtain the following:
x = b + c
x
This equation expresses x as a “function of itself”; we can take this function and plug it into the
right hand side:
x = b + c
b + c
x
and we can repeat this same procedure indefinitely:
c
x = b +
c
b + c
b +
b +
c
b + . . .
This representation of the solution of the degree two equation is known as continued fraction and
can be written as a recursive sequence:
a1 = c
an+1 = c
b+an
We can prove that this sequence is convergent and it converges to the solution of the equation minus
b (with this kind of representation we are missing the first b in the continued fraction). It’s easy to
8

prove that the sequence is bounded from above by c and from below by c , but the sequence is
b+c
quite clearly not monotonic (just calculate the first three members). At a closer look the sequence
turns out to be oscillating: the subsequence of the even members is increasing, and the subsequence
of the odd members is decreasing (this is left as an exercise to the reader). Since a2n and a2n+1
are bounded and monotonic they are both convergent. A priori they might converge to different
limits! We can explicitly calculate these limits as in the previous example:
and taking the limit of both the two members:
c
a2n+2 = c
b +
b + a2n
L = lim
n→+∞ a2n+2 =
c
b + c
b + L
This implies that the limit L satisfies the following equation:
L 2 + bL − c = 0
The positive solution of this equation is precisely the solution of (1) minus b. With the same method
we can prove that the limit of the odd members satisfy the same equation. Since the subsequences
of even and odd members converge to the same limit L, according to proposition 4.4 we deduce
that an converges to L.
In the special case b = c = 1, the solution to (1) is:
1
x = 1 +
1
1 +
1
1 +
1 +
1
1 + . . .
and this is another representation of the golden ratio.
8 Stolz-Cesàro
This last section is devoted to a technique, known as Stolz-Cesàro theorem, which is somehow the
equivalent of l’Hôpital for sequences.
Theorem 8.1 (Stolz-Cesàro). Let an be a sequence of real numbers and bn an increasing sequence
such that limn→+∞ bn = +∞. If the limit:
then the sequence an
bn
is also convergent to L:
The number L can also be ±∞.
an+1 − an
lim
n→+∞ bn+1 − bn
an
lim
n→+∞ bn
9
= L
= L

Proof. By definition of limit, for every ɛ > 0 there exists nɛ ∈ N such that for every n ≥ nɛ we
have:
L − ɛ < an+1 − an
bn+1 − bn
< L + ɛ
Since bn is increasing the number bn+1 − bn is always positive so we can multiply the inequality:
(L − ɛ)(bn+1 − bn) < an+1 − an < (L + ɛ)(bn+1 − bn)
If k ≥ nɛ, we can sum the three members of the inequality from nɛ up to k. Since the inequalities
hold for each one of the summands, they also hold for the summation:
(L − ɛ)
k
(bn+1 − bn) <
n=nɛ
k
(an+1 − an) < (L + ɛ)
n=nɛ
k
((bn+1 − bn))
In each one of these summations, only the first and the last element don’t cancel out, so we get:
n=nɛ
(L − ɛ)(bk+1 − bnɛ) < ak+1 − anɛ < (L + ɛ)(bk+1 − bnɛ)
If k is large enough we can assume that bk+1 is positive, since limk→+∞ bk = +∞, so we can divide
everything by bk+1:
(L − ɛ) 1 − bnɛ
+
bk+1
anɛ
bk+1
< ak+1
bk+1
< (L + ɛ) 1 − bnɛ
+
bk+1
anɛ
bk+1
The number nɛ is fixed, and bk+1 goes to +∞ when k is large; for this reason both anɛ
converge to zero. We can find kɛ ∈ N such that for all k ≥ kɛ we have:
and also:
Using these last two inequalities we get:
which proves that:
Example 8.2. Calculate the limit:
−ɛ < anɛ
− (L + ɛ)
bk+1
bnɛ
bk+1
−ɛ < anɛ
− (L − ɛ)
bk+1
bnɛ
bk+1
L − 2ɛ < ak+1
bk+1
ak+1
lim
k→+∞ bk+1
< L + 2ɛ
= L
1
lim
n→+∞
k + 2k + . . . + nk nk+1 According to Stolz-Cesàro the limit is equal to the following:
lim
n→+∞
(n + 1) k
(n + 1) k+1 = lim
− nk+1 n→+∞
k+1
i=0
< ɛ
< ɛ
(n + 1) k
ni − nk+1 k+1
i
10
= lim
n→+∞
bk+1
(n + 1) k
ni =
k
i=0
k+1
i
and bnɛ
bk+1

= lim
n→+∞
(1 + n −1 ) k
k
i=0
k+1
i
n i−k
= 1
k+1
k
= 1
k + 1
If k ≤ −1 the denominator is not an increasing sequence so the method cannot be applied. We
will prove later on with some other technique that for k = −1 we have:
Example 8.3. Calculate the limit:
lim
n→+∞ 1−1 + 2 −1 + . . . n −1 = +∞
lim
n→+∞
n k=1 1
k − log n
log n
According to Stolz-Cesàro the limit is equal to the following:
1
1
n+1 − log (n + 1) + log n
n+1 − log 1 +
lim
= lim
n→+∞ log (n + 1) − log n n→+∞
1
n
= lim
n→+∞
log
1 + 1
n
1
1+ 1
n
− n log 1 + 1
n log
1 + 1
n
n
= 0
We don’t know if the numerator is convergent or not, but this calculation shows that it cannot
diverge faster than log n. It also shows that n k=1 1
k goes to infinity as fast as log n.
11