Sequences

Definition A sequence {p_{n}} in X is a function f : N → X such that f(n) = pn ∈ X.

Definition The sequence {p_{n}} converges in X if there exists p ∈ X such that for each > 0,
there exists N ∈ N such that if n > N then d(p, pn) < . We say {p_{n}} converges to p. Or
equivalently, p is the limit of {p_{n}}.

Notation p_{n}→ p ⇐⇒ lim

n→∞{p_{n}} = p.

Definitions

(a) If {pn} does not converge to any point in X then it diverges.

(b) The range of {p_{n}} is {x ∈ X | x = p_{n} for some n}.

(c) We say {p_{n}} isbounded in X if the range of {p_{n}} is bounded in X.

True/False Questions Let {p_{n}} be a sequence in a metric space X.

(a) If pn → p and pn → p^{0}, then p = p^{0}. True.

(b) If {p_{n}} is bounded, then {p_{n}} converges. False.

(c) If p_{n} converges, then {p_{n}} is bounded. True.

(d) If pn → p, then p is a limit point of the range of {pn}. False.

(e) If p is a limit point of E ⊆ X, then there exists some sequence {p_{n}} such that p_{n} → p. True.

(f) p_{n}→ p if and only if every neighborhood of p contains all but finitely many terms in {p_{n}}.

True.

Remark “All but finitely many” is equivalent to “Almost all.”

Solution

a) Let > 0. Then there exists N_{1} ∈ N such that n > N1, d(p, p_{n}) <

2. Similarly there exists
N2 ∈ N such that n > N^{2}, d(p^{0}, pn) <

2. Then for n > max (N1, N2):

d(p, p^{0}) ≤ d(p, p_{n}) + d(p^{0}, p_{n}) < 2

2 = .

Thus d(p, p^{0}) < for all > 0. Thus d(p, p^{0}) = 0. Hence p = p^{0}.
b) Consider Ex. (4) where there is oscillation between two points.

c) Suppose p_{n} → p. Since 1 > 0 (our choice of ), there exists N ∈ N such that for n > N
implies d(p_{n}, p) < 1. Let r = max (1, d(p, p_{1}), . . . , d(p, p_{N})) + 1. Then p_{n}∈ B(p, r) for all n ∈ N.

d) Consider Ex. (3) where the sequence is simply one point.

e) p ∈ E^{0}. For all n ∈ N such that pn ∈ E such that d(p_{n}, p) < _{n}^{1} and p_{n} 6= p. There for a
sequence {p_{n}}. Let > 0. Choose N > 1

. Then for each n > N , we have d(p, p_{n}) < 1
n < 1

N < .

Thus this sequences {p_{n}} converges to p as desired.

f) (⇒) Suppose p_{n} → p. Given B(p, ), there exists N ∈ N such that when n > N, it fol-
lows that d(p, p_{n}) < , i.e. p_{n}∈ B(p, ), leaving only finitely many points, p_{1} through p_{n}possible.

(⇐) For all > 0, B(p, ) contains almost all {p_{n}}. For > 0, let
m = max{n ∈ N | p^{n}6∈ B(p, )}.

Then n > m implies pn∈ B(p, ), i.e. d(pn, p) < .

Theorem (Limit Laws) Suppose {s_{n}}, {t_{n}} are sequences in C, and lim

n→∞s_{n} = s, lim

n→∞t_{n} = t.

Then (a) lim

n→∞(s_{n}+ t_{n}) = s + t.

(b) lim

n→∞(cs_{n}) = cs for all c ∈ C.

(c) lim

n→∞(c + s_{n}) = c + s for all c ∈ C.

(d) lim

n→∞(s_{n}t_{n}) = st.

(e) lim

n→∞

1 sn

= 1

s provided s_{n}6= 0 for all n ∈ N, and s 6= 0.

Outline of the Proof (a) Let > 0. Since lim

n→∞s_{n} = s and lim

n→∞t_{n}= t, there exists N_{1}, N_{2} ∈ N such that if n ≥ N1, and
if n ≥ N_{2} then |s_{n}− s| <

2 and |t_{n}− s| <

2, respectively. This imples that if n ≥ max (N_{1}, N_{2}),
then

|(s_{n}+ t_{n}) − (s + t)| = |(s_{n}+ t_{n}) − (s + t)|

= |s_{n}− s| + |t_{n}− t|

< .

(b) Let > 0. Since lim

n→∞s_{n}= s, there exists N ∈ N such that if n ≥ N, then |sn− s| <
1 + |c|.
This imples that if n ≥ N, then |cs_{n}− cs| ≤ |c| |s_{n}− s| < |c|

1 + |c| < .

(d) Let > 0. Choose k = max(|s|, |t|, 1, ). Since lim

n→∞s_{n} = s and lim

n→∞t_{n} = t, there exists
N1, N2 ∈ N such that if n ≥ N^{1}, and if n ≥ N2 then |sn− s| <

3k and |tn− s| <

3k, respectively.

This imples that if n ≥ max (N_{1}, N_{2}), then

|s_{n}t_{n}− st| = |(s_{n}− s)(t_{n}− t) + s(t_{n}− t) + t(s_{n}− s)|

≤ ^{2}

9k^{2} + |s|

3k +|t|

3k

≤ 9k +

3+

3 since k ≥ |s|, |t|, =⇒ 1 ≥ |s|

k , |t|

k , k

≤ 9+

3+

3 since k ≥ 1

< .

(e) Let > 0. Since lim

n→∞s_{n} = s 6= 0, there exists N_{1}, N_{2} ∈ N such that
if n ≥ N_{1}, then |s_{n}− s| < 1

2|s| =⇒ |s| − |s_{n}| ≤ |s_{n}− s| < 1

2|s| =⇒ 1

2|s| ≤ |s_{n}|
if n ≥ N_{2}, then |s_{n}− s| < 1

2|s|^{2}, respectively.

This imples that if n ≥ max (N_{1}, N_{2}), then

1
s_{n} − 1

s

=

s_{n}− s
s_{n}s

< 2

|s|^{2} |s_{n}− s|

< .

Subsequences

Definition Let {pn} be a sequence in X and let {ni} be a sequence of natural numbers such
that n_{1} < n_{2} < n_{3} < . . . . Then {p_{n}_{i}} is a subsequence of {p_{n}}.

Examples

(a) Let {p_{n}} = {1, π,^{1}_{2}, π, ^{1}_{3}, π, . . . }. Notice this does not converge. But a subsequence does
converge. For example, only the π terms. As do the other terms.

(b) Notice that {^{1}_{2},^{2}_{3},^{3}_{4},^{4}_{5}, . . . } converges for 1. Notice that any subsequence converges. By
property (f ) on Page 1, almost all terms of the subsequence are contained in any neighbor-
hood of the limit.

Remarks

(a) Note that if p_{n} → p and {p_{n}_{k}} is any subsequence of {p_{n}}, then p_{n}_{k} → p.

(b) Question: Must every sequence contain a convergent subsequence?

No. Consider the sequence {1, 2, 3, . . . }, and note that there is no convergent subsequence of this sequence.

Recall that an ordered field F is said to have the least-upper-bound property if every nonempty,
bounded above subset E of F , the sup E ∈ E^{0} exists in F. For general metric spaces without the
least-upper-bound property, the following generalizations hold.

Theorem

(a) In a compact metric space X, every sequence contains a subsequence converging to a point
in X, i.e. if {p_{n}} is a sequence in a compact metric space, then there exists a subsequence
{p_{n}_{j}} of {p_{n}} converges to a point p ∈ X.

Proof The proof is simmilar to the proof of Theorem 2.41 (b) =⇒ (c) on p40.

(b) (Bolzano-Weierstrass) Every bounded sequence in R^{k}contains a convergent subsequence.

Proof Let R = {p_{n} | n ∈ N} = the range of {pn}.

Suppose R is a finite subset of R^{k}, there exists a point p ∈ R and a subsequence {n_{j} | j ∈
N} of {n | n ∈ N} such that

p_{n}_{j} = p ∀ j ∈ N
by the pigeon-hole principle. Since

j→∞lim p_{n}_{j} = p,
pnj is a convergent subsequence of {pn}.

Suppose R is a infinite subset of R^{k}, since R is bounded, there exists a k-cell I such that
R = {pn | n ∈ N} is an infinite subset of the compact set I.

Bisecting sides of I into 2^{k} k-cells {Q_{i}}^{2}_{i=1}^{k} with equal sizes such that

I =

2^{k}

[

i=1

Q_{i}.

Let I_{1} ∈ {Q_{i}}^{2}_{i=1}^{k} such that

I_{1}∩ R contains infinitely many elements and diam (I_{1}) = 1

2diam (I).

Next subdivide I1 and continue the process to obtain a sequence of k-cells such that
I ⊃ I_{1} ⊃ I_{2} ⊃ · · ·

I_{j}∩ R contains infinitely many elements for each j ∈ N
diam (I_{j}) = 1

2diam (I_{j−1}) = · · · = 1

2^{j} diam (I) → 0 as j → ∞

For each j ∈ N, since Ij contains infinitely many elements of {p_{n}}, there is a subsequence
{n_{j}} such that p_{n}_{j} ∈ I_{j}∩ R. Next, since lim

j→∞diam (I_{j}) = 0,

∞

\

j=1

I_{j} contains exactly one point,
say p ∈ R^{k}. This implies that

{p} =

∞

\

j=1

I_{j} and 0 ≤ lim

j→∞d(p_{n}_{j}, p) ≤ lim

j→∞diam (I_{j}) = 0 =⇒ lim

j→∞p_{n}_{j} = p.

Theorem Let X be a metric space and let {p_{n}} be a sequence in X. Suppose that E = {p_{n} |
n ∈ N} is the set of points determined by {pn} and E^{∗} is the set of all subsequential limits of
{pn} given by

E^{∗} = {p ∈ X | ∃ a subsequence {p_{n}_{i}} ⊂ {p_{n}} such that lim

i→∞p_{n}_{i} = p}.

Then E^{∗} is closed, i.e. E^{∗}^{0}

⊆ E^{∗}.

Proof

Suppose E = {q_{1}, . . . , q_{m}} contains finitely many points of X.

=⇒ E^{∗} contains finite number of points

=⇒ E^{∗}^{0}

= ∅ ⊆ E^{∗}

=⇒ E^{∗} is closed.

Suppose E contains infinitely many points of X and assume that E^{∗}0

6= ∅.

=⇒ Given q ∈ E^{∗}0

, let p_{n}_{1} ∈ {p_{n}} such that p_{n}_{1} 6= q

=⇒ δ = d(q, p_{n}_{1}) > 0.

Since q ∈ E^{∗}0

=⇒ ∃ x_{1} ∈ N_{δ/2}(q) ∩ E^{∗}\ {q} and since x_{1} ∈ E^{∗}

=⇒ ∃ n_{2} > n_{1} such that p_{n}_{2} ∈ N_{δ/2}(x_{1}) ∩ E.

=⇒ d(q, p_{n}_{2}) ≤ d(q, x_{1}) + d(x_{1}, p_{n}_{2}) < δ.

Similarly, ∃ x_{2} ∈ N_{δ/2}^{2}(q) ∩ E^{∗}\ {q} and since x_{2} ∈ E^{∗}

=⇒ ∃ n3 > n2 > n1 such that pn3 ∈ N_{δ/2}^{2}(x2) ∩ E.

=⇒ d(q, p_{n}_{3}) ≤ d(q, x_{2}) + d(x_{2}, p_{n}_{3}) < δ/2.

q
p_{n}_{1}

x_{1} p_{n}_{2}
δ

δ/2

δ/2

x_{2}
p_{n}_{3}

δ/4 δ/4

For each k ≥ 2, suppose p_{n}_{1}, . . . , p_{n}_{k} are chosen such that d(p_{n}_{j}, q) < δ/2^{j−2} for each 2 ≤ j ≤ k,
then we choose

x_{k} ∈ N_{δ/2}^{k}(q) ∩ E^{∗}\ {q} and p_{n}_{k+1} ∈ N_{δ/2}^{k}(x_{k}) ∩ E

such that

d(q, p_{n}_{k+1}) ≤ d(q, x_{k}) + d(x_{k}, p_{n}_{k+1}) < δ/2^{k−1}.

pn_{k}

xq_{k}

p_{n}_{k+1}
δ/2^{k−2}

δ/2^{k−1}

δ/2^{k}

δ/2^{k}

Since lim

k→∞d(q, p_{n}_{k}) ≤ lim

k→∞δ/2^{k−2} = 0, {p_{n}_{k}} is a subsequence of {p_{n}} converging to q and hence
q ∈ E^{∗}. Since q is an arbitrary point from q ∈ E^{∗}0

, we have shown E^{∗}0

⊆ E^{∗} and E^{∗} is closed.

Cauchy Sequences

Question How to tell if {p_{n}} converges if you don’t know the limit already?

Definition The sequence {pn} is a Cauchy sequence if for each > 0, there exists N ∈ N such
that if m, n > N then it is implied that d(p_{m}, p_{n}) < .

Definition Let E be a subset of a metric space X = (X, d). The diameter of E is defined by diam E = sup{d(p, q) | p, q ∈ E}.

Remarks

(a) Note that 0 ≤ diam E ≤ ∞.

(b) Let {p_{n}} be a sequence in X and let E_{N} = {p_{n} | n ≥ N }. Then {p_{n}} is a Cauchy sequence
if and only if

lim

N →∞diam E_{N} = 0.

(c) Let E be a subset of a metric space X = (X, d). Then diam ¯E = diam E.

Proof Since E ⊆ ¯E, we have diam E ≤ diam ¯E.

Given > 0 and for any p, q ∈ ¯E, since ¯E = E ∪ E^{0}, there exist p^{0}, q^{0} ∈ E such that
d(p, p^{0}) <

2 and d(q, q^{0}) <
2.

This implies that

d(p, q) ≤ d(p, p^{0}) + d(p^{0}, q^{0}) + d(q, q^{0}) ≤

2 + diam E +

2 = diam E + ∀ p, q ∈ ¯E, and

diam ¯E ≤ diam E + =⇒ diam ¯E ≤ diam E by letting → 0^{+}.
Hence we have diam ¯E = diam E.

(d) If K_{n} is a sequence of compact sets in X such that K_{n} ⊃ K_{n+1} for all n ∈ N and if

n→∞lim diam K_{n}= 0,

then

∞

\

n=1

K_{n} consists of exactly one point.

Theorem

(a) In any metric space X, every convergent sequence is a Cauchy sequence, i.e. if {p_{n}} ⊂ X
converges, then it is Cauchy.

Proof Suppose that lim

n→∞p_{n}= p. Then ∀ > 0 there exists N such that n > N d(p_{n}, p) <
2.
So for m, n > N we know

d(p_{n}, p_{m}) ≤ d(p_{n}, p) + d(p_{m}, p)

< 2 +

2

= .

Remark Not every Cauchy sequence converges. To see this, let X = Q. Let pn be the smallest such that m

n > π. Then {p_{n}} is Cauchy, but does not converge in Q (in R converge
to π).

(b) If X is a compact metric space and if {p_{n}} is a Cauchy sequence in X, then {p_{n}} converges
to some point of X.

Proof Consider the set S defined by

S = {p_{n}| n ∈ N}, i.e. the range set of the map p : N → X defined by p(n) = pn.
Case 1 S contains finitely many points, say S = {p_{n}_{1}, . . . , p_{n}_{m}}.

Let

δ = min{d(p_{n}_{i}, p_{n}_{j}) | 1 ≤ i < j ≤ m}.

Given > 0 such that < δ, since {p_{k}} is Cauchy, there exists N ∈ N such that
if k, l ≥ N then d(p_{k}, p_{l}) < =⇒ if k, l ≥ N then d(p_{k}, p_{l}) < δ.

Suppose ∃ k, l ≥ N such that p_{k} 6= p_{l}, then we must have

δ ≤ d(p_{k}, p_{l}) < δ =⇒ δ < δ which is a contradiction.

Hence we have

∀ k, l ≥ N, p_{k} = p_{l} =⇒ if k ≥ N, then p_{k} = p_{n}_{j} for some 1 ≤ j ≤ m.

Therefore we have

lim

k→∞p_{k} = p_{n}_{j} for some 1 ≤ j ≤ m,
i.e. {p_{k}} converges to p_{n}_{j} ∈ X.

Case 2 S contains infinitely many points.

Since X is compact and S is an infinite subset of X,

X is closed, S^{0} 6= ∅ and S^{0} ⊂ X^{0} = X =⇒ ∃ p ∈ S^{0} ⊂ X.

This implies that for each k ∈ N, there exists pnk ∈ B_{1/k}(p) ∩ S \ {p}. Thus we obtain a
subsequence {p_{n}_{k}} of {p_{n}} converging to p. This imples that

∀ > 0, ∃ N_{1} ∈ N such that if k ≥ N1, then d(p_{n}_{k}, p) < .

Also since {p_{n}} is a Cauchy sequence,

with the given > 0, ∃ N_{2} ∈ N such that if n, m ≥ N2, then d(p_{n}, p_{m}) < .

Let N = max{N_{1}, N_{2}}. Since

d(p_{n}, p) ≤ d(p_{n}, p_{n}_{n}) + d(p_{n}_{n}, p) < 2 ∀ n ≥ N =⇒ lim

n→∞p_{n} = p,
i.e. {p_{n}} converges to p ∈ X.

(c) In R^{k}, every Cauchy sequence converges.

Definition A metric space X is complete if every Cauchy sequence converges to a point in X.

Remarks

(a) Compact metric spaces are complete.

Proof Let {x_{i}} be Cauchy in X. Then {x_{i}} has a convergent subsequence. So there exist
{x_{n}_{k}} converging to x ∈ X. Fix > 0. Cauchy implies there exists N ∈ N such that
i, j > N then d(x_{i}, x_{j}) < _{2}^{}. By convergence of {x_{n}_{k}}, there exists N^{0} ∈ N such that
whenever n_{k}> N^{0}, then d(x, x_{n}_{k}) < _{2}^{}. Let N^{00} = max (N, N^{0}). Then i, n_{k}> N^{00} implies

d(x, x_{i}) ≤ d(x, x_{n}_{k}) + d(x_{n}_{k}, x_{i})

< 2+

2

= .

Hence the sequence itself converges to x and therefore X is complete.

(b) R^{k} is complete but Q is not.

(c) Let X be the ordered field R. Then X has the leat-upper-bound property if and only if every Cauchy sequence converges to a point in X. Thusan ordered field X is complete if it has the least-upper-bound property.

Proof

(=⇒) Let S be a nonempty, bounded from above subset of X = R. Choose a sufficiently large M > 0 such that

M is an upper bound of S and [−M, M ] ∩ S 6= ∅.

Let I_{1} = [−M, M ]. Divide I_{1} into 2 equal length subintervals and let I_{2} be the closed
subinterval such that

the right endpoint of I_{2} is an upper bound of S and I_{2}∩ S 6= ∅.

Continuing this process, we obtain a sequence of closed intervals {I_{n}} such that
the right endpoint of I_{n} is an upper bound of S, I_{n}∩ S 6= ∅ ∀ n ≥ 1.

and

I_{1} ⊃ I_{2} ⊃ · · · ⊃ I_{n}⊃ · · · ; lim

n→∞|I_{n}| = lim

n→∞

M
2^{n−2} = 0,
For each n ∈ N, since I^{n}∩ S 6= ∅, let xn be a point in In∩ S.

Since lim

n→∞|I_{n}| = lim

n→∞

M

2^{n−2} = 0, {x_{n}} is a Cauchy sequence and hence it converges to a
point, say x, in X = R.

(⇐=) Let {x_{n}} be a Cauchy sequence in X = R. Since {xn} is bounded, there exists
x ∈ X = R and a closed interval I = [−M, M] such that

x = sup{x_{n} | n ∈ N} and {xn} ⊂ I.

Since I is compact and {x_{n}} is a Cauchy sequence, we have

n→∞lim x_{n} = x.

Theorem Every metric space (X, d) has a completion (X^{?}, ∆). In other words, (X^{?}, ∆) is a
complete metric space containing in X.

Let X^{?} = {Cauchy sequences in X}/ ∼. We will say:

{p_{n}} ∼ {p^{0}_{n}} ⇐⇒ lim

n→∞d(p_{n}, p^{0}_{n}) = 0.

Let P, Q ∈ X^{?}. Then P = [{p_{n}}], and Q = [{p^{0}_{n}}]. We define:

∆(P, Q) = lim

n→∞d(p_{n}, q_{n}).

We claim (X^{?}, ∆) is a complete metric space and (X, d) is isometrically embedded in (X^{?}, ∆).

In other words, there is an injection i : X ,→ X^{?} such that d(p, q) = ∆(i(p), i(q)).

Remarks

(a) This is the other construction of R. But Dedekind cuts are more hardcore.

(b) Example If X = Q, then X^{?} = R. In particular, X^{?} is isometrically isomorphic to R. In
other there is a distance preserving bijection.

Definition A sequence {s_{n}} in R is monotonically increasing if s_{n} ≤ s_{n+1} for all n. Similarly
{sn} ismonotonically decreasing if sn ≥ sn+1.

Theorem Suppose {s_{n}} is a monotonic sequence of real numbers. Then {s_{n}} converges if and
only if it is bounded.

Proof

(⇐=) Suppose s_{n} ≤ s_{n+1} (the proof is analogous in the other case).

Let

E = {s_{n}| n ∈ N} = the range of {sn}.

If {s_{n}} is bounded, let

s = sup E = sup{s_{n}| n ∈ N}.

Then

sn ≤ s ∀ n ∈ N.

For every > 0, since s − < s and there is an integer N such that s − < sN ≤ s,

for otherwise s − would be an least-upper-bound of E. Since s_{n+1}≥ s_{n} for all n ∈ N, therefore
implies

if n ≥ N, then s − < s_{N} ≤ s_{n} ≤,
which shows that {s_{n}} converges to s.

(=⇒) Suppose that s = lim

n→∞sn. By taking = 1, there exists K ∈ N such that if n ≥ K, then |sn| − |s| ≤ |sn− n| < = 1 =⇒ |sn| < 1 + |s| ∀n ≥ K.

Thus

if L = max{|s_{1}|, . . . , |s_{K−1}|, 1 + |s|} =⇒ |s_{n}| ≤ L ∀ n ∈ N.

This proves that {s_{n}} is bounded.

Remarks

(a) Monotone Sequence Property Let F be an ordered field. We say F has the monotone sequence property if every bounded above monotonically increasing sequence converges.

(b) Let X be the ordered field R. Then X has the least-upper-bound property if and only if X has the monotone sequence property. Thus an ordered field X is complete if it has the monotone sequence property.

Proof

(⇐=) Let S be a nonempty, bounded from above subset of X = R. Choose a sufficiently large M > 0 such that

M is an upper bound of S and [−M, M ] ∩ S 6= ∅.

Let I_{1} = [−M, M ]. Divide I_{1} into 2 equal length subintervals and let I_{2} be the closed
subinterval such that

the right endpoint of I_{2} is an upper bound of S and I_{2}∩ S 6= ∅.

Continuing this process, we obtain a sequence of closed intervals {I_{n}} such that
the right endpoint of I_{n} is an upper bound of S, I_{n}∩ S 6= ∅ ∀ n ≥ 1.

and

I_{1} ⊃ I_{2} ⊃ · · · ⊃ I_{n}⊃ · · · ; lim

n→∞|I_{n}| = lim

n→∞

M
2^{n−2} = 0,

For each n ∈ N, by letting xn be the left endpoint of I_{n}, we obtain a bounded above
monotonically increasing sequence {xn}. Hence there exists x ∈ X = R such that

n→∞lim xn = x.

Since lim

n→∞|I_{n}| = lim

n→∞

M

2^{n−2} = 0 and, for each n ≥ 1, the right endpoint of I_{n} is an upper
bound of S, x = sup{x_{n}| n ∈ N} ∈ X = R.

(=⇒) Let {x_{n}} be a bounded above monotonically increasing sequence in X = R. Since
{x_{n}} is bounded above monotonically increasing, there exists x ∈ X = R such that

n→∞lim x_{n} = x.

Upper and Lower Limits

Definition Let {s_{n}} be a sequence of real numbers.

(a) lim

n→∞s_{n} = ∞ iff ∀ M ∈ R, ∃ N ∈ N such that if n ≥ N then sn ≥ M.

(b) lim

n→∞s_{n} = −∞ iff ∀ M ∈ R, ∃ N ∈ N such that if n ≥ N then sn ≤ M.

Definition 3.16 Let {s_{n}} be a sequence of real numbers and let E be the set of subsequential
limits of {s_{n}}defined by

E =n

x ∈ (−∞, ∞) ∪ {±∞} | ∃ {s_{n}_{k}} ⊂ {s_{n}} such that lim

k→∞s_{n}_{k} = xo
.
Let

lim sup

n→∞

s_{n} = s^{∗} = sup E,
lim inf

n→∞ sn = s∗ = inf E.

The numbers s^{∗}, s_{∗} are called the upper and lower limits of {s_{n}}; we use the notation
lim sup

n→∞

sn = s^{∗}, lim inf

n→∞ sn= s∗.

Theorem 3.17 Let {sn} be a sequence of real numbers and let E be the set of subsequential
limits of {s_{n}}. Then s^{∗} = sup E and s∗ = inf E have the following properties:

(a) s^{∗}, s∗ ∈ E, i.e. there exist subsequences {s_{n}_{k}}, {s_{m}_{j}} of {s_{n}} such that
lim

k→∞s_{n}_{k} = s^{∗}, lim

j→∞s_{m}_{j} = s∗.
Proof

If s^{∗} = ∞, then E is not bounded above; hence {s_{n}} is not bounded above, and there is a
subsequence {s_{n}_{k}} such that lim

k→∞s_{n}_{k} = ∞.

If s^{∗} ∈ R = (−∞, ∞), then E is bounded above; and at least one subsequential limit exists,
so that (a) follows from the Weierstrass Theorem.

If s^{∗} = −∞, then E contains only one element, namely −∞, and there is no subsequential
limit. Hence for any M ∈ R, there exist at most a finite number of values of n such that
s_{n} > M which implies that lim

n→∞= −∞.

(b) If x > s^{∗} ≥ s∗ > y, there exists an N ∈ N such that n ≥ N implies x > sn > y, i.e. for each

> 0, there exists an N ∈ N such that

if n ≥ N then s∗− < s_{n}< s^{∗}+ .

s^{∗}+
s∗ −

Proof For each > 0, if there are infinitely many s_{n}’s lying in (−∞, s∗− ] ∪ [s^{∗}+ , ∞),
then there exists a subsequence {s_{n}_{k}} lying completely in (−∞, s∗− ] or [s^{∗}+ , ∞).

This implies that there exists a subsequential limit x ∈ (−∞, s∗− ] ∪ [s^{∗}+ , ∞) ∪ {±∞}

which is a contradiction to the definition of s∗ = inf E or that of s^{∗} = sup E.

Hence there are only finite number of x_{n}’s lying in (−∞, s∗− ] ∪ [s^{∗}+ , ∞).

Remarks

(a) If k ≥ l ∈ N, then {sm | m ≥ k} ⊆ {s_{m} | m ≥ l} and

inf{s_{m} | m ≥ l} ≤ inf{s_{m} | m ≥ k} ≤ sup{s_{m} | m ≥ k} ≤ sup{s_{m} | m ≥ l},

i.e. sup{s_{m} | m ≥ k} is monotone decreasing in k, and inf{s_{m} | m ≥ k} is monotone
increasing in k.

(b) If s^{∗}, s∗ ∈ E, and {s_{n}_{k}}, {s_{m}_{j}} are subsequences of {s_{n}} such that

k→∞lim s_{n}_{k} = s^{∗}, lim

j→∞s_{m}_{j} = s∗ =⇒ lim inf

k→∞ s_{n}_{k} = s^{∗} = lim sup

k→∞

s_{n}_{k}, lim inf

j→∞ s_{m}_{j} = s∗ = lim sup

j→∞

s_{m}_{j}.
For each > 0, since there only exist finite number of n ∈ N such that sn > s^{∗} + or
s_{n} < s∗− , hence we have

s∗− ≤ inf{s_{m} | m ≥ n} ≤ sup{s_{m} | m ≥ n} ≤ s^{∗}+ for sufficiently large n

=⇒ s∗− ≤ lim

n→∞inf{s_{m} | m ≥ n} ≤ lim

n→∞sup{s_{m} | m ≥ n} ≤ s^{∗}+ ∀ > 0

=⇒ lim

→0^{+}(s∗− ) ≤ lim

n→∞inf{s_{m} | m ≥ n} ≤ lim

n→∞sup{s_{m} | m ≥ n} ≤ lim

→0^{+}(s^{∗}+ )

=⇒ s∗ ≤ lim

n→∞inf{s_{m} | m ≥ n} ≤ lim

n→∞sup{s_{m} | m ≥ n} ≤ s^{∗}.

On the other hand for each l ∈ N, since {smj | j ≥ l} and {s_{n}_{k} | k ≥ l} are subsequences of
{sn | n ≥ ml} and {sm | m ≥ nl} respectively, we have

inf{sn| n ≥ ml} ≤ smj and sn_{k} ≤ sup{sm | m ≥ nl} ∀ j, k ≥ l

=⇒ inf{s_{n}| n ≥ m_{l}} ≤ lim

j→∞s_{m}_{j} = s∗ and s^{∗} = lim

k→∞s_{n}_{k} ≤ sup{s_{m} | m ≥ n_{l}} ∀ l ∈ N

=⇒ inf{s_{n}| n ≥ m_{l}} ≤ s∗ ≤ s^{∗} ≤ sup{s_{m} | m ≥ n_{l}} ∀ l ∈ N

=⇒ lim

l→∞inf{s_{n} | n ≥ m_{l}} ≤ s_{∗} ≤ s^{∗} ≤ lim

l→∞sup{s_{m} | m ≥ n_{l}}

=⇒ lim

n→∞inf{s_{m} | m ≥ n} ≤ s∗ ≤ s^{∗} ≤ lim

n→∞sup{s_{m} | m ≥ n}.

Hence

lim sup

n→∞

s_{n}= s^{∗} = lim

n→∞sup{s_{m} | m ≥ n} = inf{sup{s_{m} | m ≥ n} | n ∈ N},
lim inf

n→∞ s_{n}= s_{∗} = lim

n→∞inf{s_{m} | m ≥ n} = sup{inf{s_{m}| m ≥ n} | n ∈ N}.

(c) If x > s^{∗} = lim

n→∞sup{sm | m ≥ n} and lim

n→∞inf{sm | m ≥ n} = s∗ > y, there exists an
N ∈ N depending on x − s^{∗} > 0 and s∗ − y > 0 such that if n ≥ N then

max{| sup{s_{m} | m ≥ n} − s^{∗}|, |s∗− inf{s_{m} | m ≥ n}|} < min{x − s^{∗}, s∗− y} ∀ n ≥ N

=⇒ sup{s_{m} | m ≥ n} − s^{∗} < x − s^{∗} and s∗− inf{s_{m} | m ≥ n} < s∗− y ∀ n ≥ N

=⇒ s_{n}≤ sup{s_{m} | m ≥ n} < x and s_{n}≥ inf{s_{m} | m ≥ n} > y ∀ n ≥ N

=⇒ y < inf{s_{m} | m ≥ n} ≤ s_{n}≤ sup{s_{m} | m ≥ n} < x ∀ n ≥ N.

Examples

(a) Let {s_{n}} = Q, be a sequence containing all rationals. Then every real number is a subse-
quential limit, and

lim sup

n→∞

s_{n}= ∞, lim inf

n→∞ s_{n} = −∞,
i.e. the set of subsequential limits of {s_{n}} is E = R ∪ {±∞}.

(b) Let s_{n}= (−1)^{n}

1 + (1/n). Then

lim sup

n→∞

s_{n} = 1, lim inf

n→∞ s_{n} = −1.

(c) For a real-valued sequence {s_{n}},

n→∞lim s_{n} = s ⇐⇒ lim sup

n→∞

s_{n} = lim inf

n→∞ s_{n} = s.

Theorem If s_{n} ≤ t_{n} for n ≥ N, where N is fixed, then
lim inf

n→∞ s_{n}≤ lim inf

n→∞ t_{n}, lim sup

n→∞

s_{n} ≤ lim sup

n→∞

t_{n}.

Some Special Sequences

Theorem

(a) If p > 0, then lim

n→∞

1
n^{p} = 0.

Proof For each > 0, there exists n ∈ N such that n > 1

^{1/p} ⇐⇒ n^{p} > 1

⇐⇒ > 1
n^{p} = | 1

n^{p} − 0|.

(b) p > 0, then lim

n→∞

√n

p = 1.

Proof If p > 1, by letting x_{n} = √^{n}

p − 1 > 0, then
1 + n x_{n}≤ (1 + x_{n})^{n}= p =⇒ x_{n} ≤ p − 1

n =⇒ lim

n→∞x_{n} = 0 ⇐⇒ lim

n→∞

√n

p = 1.

If p < 1, by letting p = 1

q for some q > 1, we have

n→∞lim

√n

q = 1 =⇒ lim

n→∞

1

√n

q = 1 =⇒ lim

n→∞

√n

p = lim

n→∞

1

√n

q = 1.

(c) lim

n→∞

√n

n = 1.

Proof Let x_{n} = √^{n}

n − 1. Then x_{n}≥ 0 for each n ∈ N and
n = (1 + x_{n})^{n}≥ n(n − 1)

2 x^{2}_{n} =⇒ 0 ≤ x_{n}≤

r 2

n − 1 for n ≥ 2 =⇒ lim

n→∞x_{n} = 0.

(d) If p > 0 and α ∈ R, then lim

n→∞

n^{α}

(1 + p)^{n} = 0.

Proof Let k ∈ N such that k > α. If n ∈ N and n

2 > k, then
(1 + p)^{n}>n

k

p^{k} = n(n − 1) · · · (n − k + 1)

k! p^{k}> n^{k}p^{k}

2^{k}k! =⇒ 0 < n^{α}

(1 + p)^{n} < 2^{k}k!

p^{k} n^{α−k}.
(e) If |x| < 1, then lim

n→∞x^{n}= 0.

Proof Take α = 0 and p > 0 such that |x| = 1

1 + p as in (d).

Series

Question What does this mean?

1 + 1 4+ 1

9+ 1 16+ 1

25+ · · · = π^{2}
6 .

Remark Sum of natural numbers to negative even powers always has a nice form.

Consider also:

1 − 1 + 1 − 1 + 1 − 1 + . . .

You can associate these differently and get different limits. This tells us we cannot assume associativity in infinite sums.

Notation Let {a_{n}} be a real sequence. Then:

j

X

n=i

a_{n} = a_{i}+ a_{i+1}+ · · · + a_{j},
when i < j ∈ N.

Definition The nth partial sum of {a_{k}} is
s_{n}=

n

X

k=1

a_{k}.

Remark {s_{n}} is a real sequence. Sometimes {s_{n}} is called an infinite series.

Definition If s_{n} → s we write

∞

X

n=1

a_{n} = lim

n→∞

n

X

k=1

a_{k} = s

Question When does an infinite series converge? When its sequence of partial sums converge.

Example Does

∞

X

n=1

1

n converge?

Consider partial sums where s_{n}=

n

X

k=1

1

k. We use the Cauchy criterion. For m < n, then d(sm, sn) = sn− sm

= s_{m+1}+ s_{m+2}+ · · · + s_{n}

= 1

m + 1 + 1

m + 2 + · · · + 1

n ≥ n − m n .

The inequality comes from observing that all the terms in the sum are less than or equal to 1
n.
Thus s_{2n} − s_{n} ≥ 2n − n

2n = 1

2. Therefore this sequence is not Cauchy. Hence {s_{n}} does not
converge.

Cauchy Criterion for series X

an converges if and only if for all > 0 there exists N ∈ N such that m ≥ n > N implies

m

X

k=n

a_{k}

< .

Corollary (Divergence Test) If X

a_{n} converges, then lim

n→∞a_{n} = 0.

Remarks (a) X 1

n is called the harmonic series.

(b) The corollary’s converse is not true (counter-example is harmonic series).

Theorem If a_{n} ≥ 0, then X

a_{n} converges if and only if partial sums {s_{n}} form a bounded
sequence.

Proof Since a_{k} ≥ 0 for all k ∈ N, {sn} is monotonically increasing. Thus {s_{n}} converges, i.e.

Xan converges, if and only if {sn} is bounded.

Comparison Test 1. If X

c_{n} converges and |a_{n}| ≤ c_{n} for almost all n, then X

a_{n} converges.

2. If X

d_{n} diverges to +∞ and a_{n} ≥ d_{n} for almost all n, then X

a_{n} diverges as well.

Proof

1. Let > 0. Since X

c_{n} converges, it satisfies Cauchy Criterion. Thus there exists N ∈ N
such that m ≥ n ≥ N implies:

m

X

k=n

c_{k} ≤

m

X

k=n

c_{k}

< .

Thus

m

X

k=n

a_{k}

≤

m

X

k=n

|a_{k}| ≤

m

X

k=n

c_{k}.

This follows from the assumption that |a_{n}| ≤ c_{n} for almost all n so let N be at least larger
than the last n for which c_{n} < |a_{n}|. The resulting inequality satisfies the Cauchy Criterion
and thusX

a_{n} converges.

2. Follows from (a) via contrapositive. (Also, partial sums form a bounded sequence.)

Geometric Series If |x| < 1, then

∞

X

n=0

x^{n}= 1
1 − x.
If |x| ≥ 1, then X

x^{n} diverges.

Proof If x 6= 1, let sn =

n

X

k=0

xk = 1 + x + · · · + x^{n}. Then sn = 1 − x^{n+1}

1 − x by multiplying both sides by 1 − x. Thus it follows:

n→∞lim s_{n} = 1

1 − x if |x| < 1

If |x| > 1, then {s_{n}} does not converge. Similarly if x = ±1, use the divergence test to verify
{s_{n}} does not converge.

Example

∞

X

n=0

1

n! converges. Notice

s_{n}= 1 + 1 + 1
2! + 1

3! + 1

4! + · · · + 1 n!

≤ 1 + 1 + 1 2+ 1

2^{2} + 1

2^{3} + · · · + 1
2^{n−1}

< 3.

Thus it is bounded and since each term is nonnegative, it is monotonically increasing. Thus it converges.

Definition e =

∞

X

n=0

1 n!. Theorem e = lim

n→∞(1 + 1
n)^{n}.
Proof

Let s_{n}=

n

X

k=0

1

k!, t_{n} = 1 + 1
n

n

. By the binomial theorem,

tn= 1 + 1 + 1

2! 1 − 1 n + 1

3! 1 − 1 n

1 − 2

n + · · · + 1

n! 1 − 1 n

1 − 2

n · · · 1 − n − 1 n .

Hence t_{n}≤ s_{n} ≤ e for all n ∈ N, so that

lim sup

n→∞

tn≤ e.

Next, if n ≥ m, then

t_{n}≥ 1 + 1 + 1

2! 1 − 1

n + · · · + 1

m! 1 − 1 n

1 − 2

n · · · 1 − m − 1 n .

By keeping m fixed and letting n → ∞ on both sides, we get lim inf

n→∞ t_{n} ≥ 1 + 1 + 1

2! + · · · + 1

m! =⇒ lim inf

n→∞ t_{n} ≥ s_{m} ∀ m ∈ N.

Letting m → ∞, we finally get e ≤ lim inf

n→∞ t_{n} ≤ lim sup

n→∞

t_{n}≤ e =⇒ e = lim

n→∞t_{n} = lim

n→∞(1 + 1
n)^{n}.

Remark

e − s_{n} = 1

(n + 1)! + 1

(n + 2)! + . . .

< 1

(n + 1)!(1 + 1

n + 1 + 1

(n + 1)^{2} + . . . )

= 1

(n + 1)!

1

1 − 1/(n + 1)

= 1 n! n

Theorem e 6∈ Q.

Proof Suppose e = m

n for m, n > 0. Then

0 < n! (e − s_{n})

| {z }

∈N

< 1 n < 1

Recall A series converges if its partial sums converges.

Example 1 + 1 2+ 1

4+ 1 8+ 1

16+ . . . converges because partial sums converge.

Cauchy’s Theorem If a_{1} ≥ a_{2} ≥ a_{2}· · · ≥ 0, i.e. monotonically decreasing, then

∞

X

n=1

a_{n} con-

verges if and only if

∞

X

k=0

2^{k}a_{2}^{k} = a_{1}+ 2a_{2}+ 4a_{4} + 8a_{8}+ . . . converges.

Proof Compare s_{n} = a_{1} + · · · + a_{n} and t_{k} = a_{1} + 2a_{2} + · · · + 2^{k}a_{2}^{k}. Consider the following
grouping of the finite sum:

s_{n} = (a_{1}) + (a_{2}+ a_{3}) + (a_{4} + · · · + a_{7}) + · · · + a_{n}

t_{k} = (a_{1}) + (a_{2}+ a_{2}) + (a_{4} + · · · + a_{4}) + · · · + (a_{2}^{k} + · · · + a_{2}^{k})

If t_{k} converges, since s_{n} < t_{k} ∀ n < 2^{k} and ∀ k ∈ N,

=⇒ sn≤ lim

k→∞tk ∀ n ∈ N

=⇒ s_{n} converges since s_{n} is monotonically increasing and bounded above.

On the other hand,

if s_{n} converges, since 2s_{n}> t_{k}− a_{1} ∀ n ≥ 3 and ∀ 2^{k} < n,

=⇒ 2 lim

n→∞s_{n} ≥ t_{k}− a_{1} ∀ k ∈ N

=⇒ t_{k} converges since t_{k}− a_{1} is monotonically increasing and bounded above.

This proves that both series diverge or converge simultaneously.

Theorem Consider P 1

n^{p}. Claim is that this converges if p > 1 and diverges if p ≤ 1.

Proof If p ≤ 0, terms do not go to zero, so the series diverges. If p > 0, look at:

X

k

2^{k} 1

(2^{k})^{p} =X

k

2^{(1−p)k},

which is geometric. Thus it converges if and only if 2^{1−p}< 1. This only happens when 1 − p < 0
and hence p > 1.

Remark We were able to turn a harmonic-like series into a geometric-like series.

Theorem (Root Test) Given P an, let α = lim

n→∞

p|an _{n}|. Then
(a) if α < 1, then P a_{n} converges;

(b) if α > 1, then P a_{n} diverges;

(c) if α = 1, then the test is inconclusive.

Proof If α < 1, we can choose β so that α < β < 1, and an integer N such that
p|an _{n}| < β for n ≥ N, by Theorem 3.17(b).

That is, if n ≥ N, then |an| < β^{n}. Since 0 < β < 1, P β^{n} converges. Therefore, P an converges
(absolutely) by the comparison test.

If α > 1, then, again by Theorem 3.17(b), there exists a sequence {n_{k}} such that

nk

q

|a_{n}_{k}| → α > 1.

Hence |a_{n}| > 1 for infinitely many values of n, and

n→∞lim a_{n}6= 0 =⇒ X

a_{n} diverges.

Since P 1

n diverges, P 1

n^{2} converges and α = 1 in both cases, the test is inconclusive if α = 1.

Theorem (Ratio Test) The series P an

(a) converges if lim sup

an+1

a_{n}
< 1,
(b) diverges if

a_{n+1}

a_{n}

≥ 1 for n ≥ n_{0}, where n_{0} is some fixed integer.

Proof If lim sup

a_{n+1}
an

< 1, there exist β < 1 and an integer N such that

a_{n+1}

a_{n}

< β < 1 for n ≥ N.

This implies that

|a_{N +k}| < β|a_{N +k−1}| < β^{2}|a_{N +k−2}| < · · · < β^{k}|a_{n}| for each k ∈ N.

Therefore,P an converges (absolutely) by the comparison test.

If |a_{n+1}| ≥ |a_{n}| for n ≥ n_{0}, then

n→∞lim an6= 0 =⇒ X

an diverges.

Definition A power series is of the form

∞

X

k=0

c_{n}z^{n} where c_{n} ∈ C.

Theorem Let α = lim supp|c^{n} n|. Let r = _{α}^{1}. Then the power series converges if |z| < R and
diverges |z| > R. We call R the radius of convergence.

Proof Use the root test so consider lim sup

n→∞

p|cn _{n}z^{n}| = lim sup

n→∞

|z|p|c^{n} _{n}| = |z| lim sup

n→∞

p|cn _{n}|.

Notice that this less than 1, and thus converges, when |z| is less than 1 over the limsup.

Definition A series converges absolutely if P |an| converges.

Theorem P a_{n} converges absolutely implies P a_{n} converges.

Proof |

m

X

k=n

a_{k}| ≤

m

X

k=n

|a_{k}| < , by the Cauchy Criterion since P |a_{k}| converges.

Example If a series converges, it does not necessarily converge absolutely. Consider the series
1 − ^{1}_{2}+^{1}_{3} −^{1}_{4} + . . . , which converges by alternating series test but does not converge absolutely.

Question If the terms in a convergent series are rearranged, must it converge to same sum? Not all the time, but it does if the series converges absolutely.

Riemann’s Theorem If a series P a_{n} converges but not absolutely, then we can form a rear-
rangement that has any limsup and liminf you’d like.

Example Rearrange 1 − ^{1}_{2} + ^{1}_{3} − ^{1}_{4} + . . . to converge to 4. We want to the partial sums to
converge to 4. Consider the positive terms in sequence that sum to at least 4 (notice the positive
terms diverge). Then use a s many negative terms you need to go back, as many positive terms
you need to go forward, et cetera. These partial sums converge to 4.