0, there exists N ∈ N such that if n &gt

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 → p⁰, then p = p⁰. 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 N₁ ∈ N such that n > N1, d(p, p_n) < 

2. Similarly there exists N2 ∈ N such that n > N², d(p⁰, pn) < 

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

d(p, p⁰) ≤ d(p, p_n) + d(p⁰, p_n) < 2

2 = .

Thus d(p, p⁰) <  for all  > 0. Thus d(p, p⁰) = 0. Hence p = p⁰. 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, p₁), . . . , 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 ∈ E⁰. For all n ∈ N such that pn ∈ E such that d(p_n, p) < _n¹ 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, p₁ through p_npossible.

(⇐) For all  > 0, B(p, ) contains almost all {p_n}. For  > 0, let m = max{n ∈ N | pⁿ6∈ 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_nt_n) = st.

(e) lim

n→∞

1 sn

= 1

s provided s_n6= 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 N₁, N₂ ∈ N such that if n ≥ N1, and if n ≥ N₂ then |s_n− s| < 

2 and |t_n− s| < 

2, respectively. This imples that if n ≥ max (N₁, N₂), 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 ≥ N¹, and if n ≥ N2 then |sn− s| < 

3k and |tn− s| < 

3k, respectively.

This imples that if n ≥ max (N₁, N₂), then

|s_nt_n− st| = |(s_n− s)(t_n− t) + s(t_n− t) + t(s_n− s)|

≤ ²

9k² + |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 N₁, N₂ ∈ N such that if n ≥ N₁, then |s_n− s| < 1

2|s| =⇒ |s| − |s_n| ≤ |s_n− s| < 1

2|s| =⇒ 1

2|s| ≤ |s_n| if n ≥ N₂, then |s_n− s| < 1

2|s|², respectively.

This imples that if n ≥ max (N₁, N₂), then    

1 s_n − 1

s    

=    

s_n− s s_ns

< 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 n₁ < n₂ < n₃ < . . . . Then {p_ni} is a subsequence of {p_n}.

Examples

(a) Let {p_n} = {1, π,¹₂, π, ¹₃, 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 {¹₂,²₃,³₄,⁴₅, . . . } 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_nk} is any subsequence of {p_n}, then p_nk → 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 ∈ E⁰ 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_nj} 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^kcontains 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_nj = p ∀ j ∈ N by the pigeon-hole principle. Since

j→∞lim p_nj = 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 I₁ ∈ {Q_i}²_i=1^k such that

I₁∩ R contains infinitely many elements and diam (I₁) = 1

2diam (I).

Next subdivide I1 and continue the process to obtain a sequence of k-cells such that I ⊃ I₁ ⊃ I₂ ⊃ · · ·

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_nj ∈ 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_nj, p) ≤ lim

j→∞diam (I_j) = 0 =⇒ lim

j→∞p_nj = 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_ni} ⊂ {p_n} such that lim

i→∞p_ni = p}.

Then E^∗ is closed, i.e. E^∗
0

⊆ E^∗.

Proof

Suppose E = {q₁, . . . , 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_n1 ∈ {p_n} such that p_n1 6= q

=⇒ δ = d(q, p_n1) > 0.

Since q ∈ E^∗
0

=⇒ ∃ x₁ ∈ N_δ/2(q) ∩ E^∗\ {q} and since x₁ ∈ E^∗

=⇒ ∃ n₂ > n₁ such that p_n2 ∈ N_δ/2(x₁) ∩ E.

=⇒ d(q, p_n2) ≤ d(q, x₁) + d(x₁, p_n2) < δ.

Similarly, ∃ x₂ ∈ N_δ/2²(q) ∩ E^∗\ {q} and since x₂ ∈ E^∗

=⇒ ∃ n3 > n2 > n1 such that pn3 ∈ N_δ/2²(x2) ∩ E.

=⇒ d(q, p_n3) ≤ d(q, x₂) + d(x₂, p_n3) < δ/2.

q p_n1

x₁ p_n2 δ

δ/2

δ/2

x₂ p_n3

δ/4 δ/4

For each k ≥ 2, suppose p_n1, . . . , p_nk are chosen such that d(p_nj, q) < δ/2^j−2 for each 2 ≤ j ≤ k, then we choose

x_k ∈ N_δ/2^k(q) ∩ E^∗\ {q} and p_nk+1 ∈ N_δ/2^k(x_k) ∩ E

such that

d(q, p_nk+1) ≤ d(q, x_k) + d(x_k, p_nk+1) < δ/2^k−1.

pn_k

xq_k

p_nk+1 δ/2^k−2

δ/2^k−1

δ/2^k

δ/2^k

Since lim

k→∞d(q, p_nk) ≤ lim

k→∞δ/2^k−2 = 0, {p_nk} 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 ∪ E⁰, there exist p⁰, q⁰ ∈ E such that d(p, p⁰) < 

2 and d(q, q⁰) <  2.

This implies that

d(p, q) ≤ d(p, p⁰) + d(p⁰, q⁰) + d(q, q⁰) ≤ 

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_n1, . . . , p_nm}.

Let

δ = min{d(p_ni, p_nj) | 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_nj for some 1 ≤ j ≤ m.

Therefore we have

lim

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

Case 2 S contains infinitely many points.

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

X is closed, S⁰ 6= ∅ and S⁰ ⊂ X⁰ = X =⇒ ∃ p ∈ S⁰ ⊂ X.

This implies that for each k ∈ N, there exists pnk ∈ B_1/k(p) ∩ S \ {p}. Thus we obtain a subsequence {p_nk} of {p_n} converging to p. This imples that

∀  > 0, ∃ N₁ ∈ N such that if k ≥ N1, then d(p_nk, p) < .

Also since {p_n} is a Cauchy sequence,

with the given  > 0, ∃ N₂ ∈ N such that if n, m ≥ N2, then d(p_n, p_m) < .

Let N = max{N₁, N₂}. Since

d(p_n, p) ≤ d(p_n, p_nn) + d(p_nn, 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_nk} 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_nk}, there exists N⁰ ∈ N such that whenever n_k> N⁰, then d(x, x_nk) < ₂^. Let N⁰⁰ = max (N, N⁰). Then i, n_k> N⁰⁰ implies

d(x, x_i) ≤ d(x, x_nk) + d(x_nk, 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 I₁ = [−M, M ]. Divide I₁ into 2 equal length subintervals and let I₂ be the closed subinterval such that

the right endpoint of I₂ is an upper bound of S and I₂∩ 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

I₁ ⊃ I₂ ⊃ · · · ⊃ I_n⊃ · · · ; lim

n→∞|I_n| = lim

n→∞

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

Since lim

n→∞|I_n| = lim

n→∞

M

2ⁿ⁻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} ∼ {p⁰_n} ⇐⇒ lim

n→∞d(p_n, p⁰_n) = 0.

Let P, Q ∈ X^?. Then P = [{p_n}], and Q = [{p⁰_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{|s₁|, . . . , |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 I₁ = [−M, M ]. Divide I₁ into 2 equal length subintervals and let I₂ be the closed subinterval such that

the right endpoint of I₂ is an upper bound of S and I₂∩ 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

I₁ ⊃ I₂ ⊃ · · · ⊃ I_n⊃ · · · ; lim

n→∞|I_n| = lim

n→∞

M 2ⁿ⁻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

2ⁿ⁻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_nk} ⊂ {s_n} such that lim

k→∞s_nk = 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_nk}, {s_mj} of {s_n} such that lim

k→∞s_nk = s^∗, lim

j→∞s_mj = s∗. Proof

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

k→∞s_nk = ∞.

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_nk} 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_nk}, {s_mj} are subsequences of {s_n} such that

k→∞lim s_nk = s^∗, lim

j→∞s_mj = s∗ =⇒ lim inf

k→∞ s_nk = s^∗ = lim sup

k→∞

s_nk, lim inf

j→∞ s_mj = s∗ = lim sup

j→∞

s_mj. 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_nk | 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_mj = s∗ and s^∗ = lim

k→∞s_nk ≤ 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 x²_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^kp^k

2^kk! =⇒ 0 < n^α

(1 + p)ⁿ < 2^kk!

p^k n^α−k. (e) If |x| < 1, then lim

n→∞xⁿ= 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+ · · · = π² 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

xⁿ= 1 1 − x. If |x| ≥ 1, then X

xⁿ diverges.

Proof If x 6= 1, let sn =

n

X

k=0

xk = 1 + x + · · · + xⁿ. Then sn = 1 − xⁿ⁺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

2² + 1

2³ + · · · + 1 2ⁿ⁻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 a₁ ≥ a₂ ≥ a₂· · · ≥ 0, i.e. monotonically decreasing, then

∞

X

n=1

a_n con-

verges if and only if

∞

X

k=0

2^ka₂^k = a₁+ 2a₂+ 4a₄ + 8a₈+ . . . converges.

Proof Compare s_n = a₁ + · · · + a_n and t_k = a₁ + 2a₂ + · · · + 2^ka₂^k. Consider the following grouping of the finite sum:

s_n = (a₁) + (a₂+ a₃) + (a₄ + · · · + a₇) + · · · + a_n

t_k = (a₁) + (a₂+ a₂) + (a₄ + · · · + a₄) + · · · + (a₂^k + · · · + a₂^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− a₁ ∀ n ≥ 3 and ∀ 2^k < n,

=⇒ 2 lim

n→∞s_n ≥ t_k− a₁ ∀ k ∈ N

=⇒ t_k converges since t_k− a₁ 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| < βⁿ. Since 0 < β < 1, P βⁿ 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_nk| → α > 1.

Hence |a_n| > 1 for infinitely many values of n, and

n→∞lim a_n6= 0 =⇒ X

a_n diverges.

Since P 1

n diverges, P 1

n² 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 ≥ n₀, where n₀ 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}| < β²|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 ≥ n₀, then

n→∞lim an6= 0 =⇒ X

an diverges.

Definition A power series is of the form

∞

X

k=0

c_nzⁿ where c_n ∈ C.

Theorem Let α = lim supp|cⁿ n|. Let r = _α¹. 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 _nzⁿ| = lim sup

n→∞

|z|p|cⁿ _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 − ¹₂+¹₃ −¹₄ + . . . , 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 − ¹₂ + ¹₃ − ¹₄ + . . . 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.