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

## Full text

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Sequences

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

Definition The sequence {pn} 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 {pn} converges to p. Or equivalently, p is the limit of {pn}.

Notation pn→ p ⇐⇒ lim

n→∞{pn} = p.

Definitions

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

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

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

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

(a) If pn → p and pn → p0, then p = p0. True.

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

(c) If pn converges, then {pn} 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 {pn} such that pn → p. True.

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

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, pn) < 

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, pn) + d(p0, pn) < 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 pn → p. Since 1 > 0 (our choice of ), there exists N ∈ N such that for n > N implies d(pn, p) < 1. Let r = max (1, d(p, p1), . . . , d(p, pN)) + 1. Then pn∈ 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(pn, p) < n1 and pn 6= p. There for a sequence {pn}. Let  > 0. Choose N > 1

. Then for each n > N , we have d(p, pn) < 1 n < 1

N < .

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Thus this sequences {pn} converges to p as desired.

f) (⇒) Suppose pn → p. Given B(p, ), there exists N ∈ N such that when n > N, it fol- lows that d(p, pn) < , i.e. pn∈ B(p, ), leaving only finitely many points, p1 through pnpossible.

(⇐) For all  > 0, B(p, ) contains almost all {pn}. 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 {sn}, {tn} are sequences in C, and lim

n→∞sn = s, lim

n→∞tn = t.

Then (a) lim

n→∞(sn+ tn) = s + t.

(b) lim

n→∞(csn) = cs for all c ∈ C.

(c) lim

n→∞(c + sn) = c + s for all c ∈ C.

(d) lim

n→∞(sntn) = st.

(e) lim

n→∞

1 sn

= 1

s provided sn6= 0 for all n ∈ N, and s 6= 0.

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

n→∞sn = s and lim

n→∞tn= t, there exists N1, N2 ∈ N such that if n ≥ N1, and if n ≥ N2 then |sn− s| < 

2 and |tn− s| < 

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

|(sn+ tn) − (s + t)| = |(sn+ tn) − (s + t)|

= |sn− s| + |tn− t|

< .

(b) Let  > 0. Since lim

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

1 + |c| < .

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

n→∞sn = s and lim

n→∞tn = 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

|sntn− st| = |(sn− s)(tn− t) + s(tn− t) + t(sn− 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

< .

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(e) Let  > 0. Since lim

n→∞sn = s 6= 0, there exists N1, N2 ∈ N such that if n ≥ N1, then |sn− s| < 1

2|s| =⇒ |s| − |sn| ≤ |sn− s| < 1

2|s| =⇒ 1

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

2|s|2, respectively.

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

1 sn − 1

s

=

sn− s sns

< 2

|s|2 |sn− 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 {pni} is a subsequence of {pn}.

Examples

(a) Let {pn} = {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 pn → p and {pnk} is any subsequence of {pn}, then pnk → 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 {pn} is a sequence in a compact metric space, then there exists a subsequence {pnj} of {pn} 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 Rkcontains a convergent subsequence.

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

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

pnj = p ∀ j ∈ N by the pigeon-hole principle. Since

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

Suppose R is a infinite subset of Rk, 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 2k k-cells {Qi}2i=1k with equal sizes such that

I =

2k

[

i=1

Qi.

Let I1 ∈ {Qi}2i=1k 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 ⊃ · · ·

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

2diam (Ij−1) = · · · = 1

2j diam (I) → 0 as j → ∞

For each j ∈ N, since Ij contains infinitely many elements of {pn}, there is a subsequence {nj} such that pnj ∈ Ij∩ R. Next, since lim

j→∞diam (Ij) = 0,

\

j=1

Ij contains exactly one point, say p ∈ Rk. This implies that

{p} =

\

j=1

Ij and 0 ≤ lim

j→∞d(pnj, p) ≤ lim

j→∞diam (Ij) = 0 =⇒ lim

j→∞pnj = p.

Theorem Let X be a metric space and let {pn} be a sequence in X. Suppose that E = {pn | 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 {pni} ⊂ {pn} such that lim

i→∞pni = p}.

Then E is closed, i.e. E0

⊆ E.

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Proof

Suppose E = {q1, . . . , qm} 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 pn1 ∈ {pn} such that pn1 6= q

=⇒ δ = d(q, pn1) > 0.

Since q ∈ E0

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

=⇒ ∃ n2 > n1 such that pn2 ∈ Nδ/2(x1) ∩ E.

=⇒ d(q, pn2) ≤ d(q, x1) + d(x1, pn2) < δ.

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

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

=⇒ d(q, pn3) ≤ d(q, x2) + d(x2, pn3) < δ/2.

q pn1

x1 pn2 δ

δ/2

δ/2

x2 pn3

δ/4 δ/4

For each k ≥ 2, suppose pn1, . . . , pnk are chosen such that d(pnj, q) < δ/2j−2 for each 2 ≤ j ≤ k, then we choose

xk ∈ Nδ/2k(q) ∩ E\ {q} and pnk+1 ∈ Nδ/2k(xk) ∩ E

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such that

d(q, pnk+1) ≤ d(q, xk) + d(xk, pnk+1) < δ/2k−1.

pnk

xqk

pnk+1 δ/2k−2

δ/2k−1

δ/2k

δ/2k

Since lim

k→∞d(q, pnk) ≤ lim

k→∞δ/2k−2 = 0, {pnk} is a subsequence of {pn} 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 {pn} 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(pm, pn) < .

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 {pn} be a sequence in X and let EN = {pn | n ≥ N }. Then {pn} is a Cauchy sequence if and only if

lim

N →∞diam EN = 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.

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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 Kn is a sequence of compact sets in X such that Kn ⊃ Kn+1 for all n ∈ N and if

n→∞lim diam Kn= 0,

then

\

n=1

Kn consists of exactly one point.

Theorem

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

Proof Suppose that lim

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

d(pn, pm) ≤ d(pn, p) + d(pm, 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 {pn} is Cauchy, but does not converge in Q (in R converge to π).

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

Proof Consider the set S defined by

S = {pn| 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 = {pn1, . . . , pnm}.

Let

δ = min{d(pni, pnj) | 1 ≤ i < j ≤ m}.

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

Suppose ∃ k, l ≥ N such that pk 6= pl, then we must have

δ ≤ d(pk, pl) < δ =⇒ δ < δ which is a contradiction.

Hence we have

∀ k, l ≥ N, pk = pl =⇒ if k ≥ N, then pk = pnj for some 1 ≤ j ≤ m.

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Therefore we have

lim

k→∞pk = pnj for some 1 ≤ j ≤ m, i.e. {pk} converges to pnj ∈ 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 ∈ B1/k(p) ∩ S \ {p}. Thus we obtain a subsequence {pnk} of {pn} converging to p. This imples that

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

Also since {pn} is a Cauchy sequence,

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

Let N = max{N1, N2}. Since

d(pn, p) ≤ d(pn, pnn) + d(pnn, p) < 2 ∀ n ≥ N =⇒ lim

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

(c) In Rk, 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 {xi} be Cauchy in X. Then {xi} has a convergent subsequence. So there exist {xnk} converging to x ∈ X. Fix  > 0. Cauchy implies there exists N ∈ N such that i, j > N then d(xi, xj) < 2. By convergence of {xnk}, there exists N0 ∈ N such that whenever nk> N0, then d(x, xnk) < 2. Let N00 = max (N, N0). Then i, nk> N00 implies

d(x, xi) ≤ d(x, xnk) + d(xnk, xi)

<  2+ 

2

= .

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

(b) Rk 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= ∅.

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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 {In} such that the right endpoint of In is an upper bound of S, In∩ S 6= ∅ ∀ n ≥ 1.

and

I1 ⊃ I2 ⊃ · · · ⊃ In⊃ · · · ; lim

n→∞|In| = 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→∞|In| = lim

n→∞

M

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

(⇐=) Let {xn} 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{xn | n ∈ N} and {xn} ⊂ I.

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

n→∞lim xn = 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:

{pn} ∼ {p0n} ⇐⇒ lim

n→∞d(pn, p0n) = 0.

Let P, Q ∈ X?. Then P = [{pn}], and Q = [{p0n}]. We define:

∆(P, Q) = lim

n→∞d(pn, qn).

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

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

Proof

(⇐=) Suppose sn ≤ sn+1 (the proof is analogous in the other case).

Let

E = {sn| n ∈ N} = the range of {sn}.

If {sn} is bounded, let

s = sup E = sup{sn| 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 sn+1≥ sn for all n ∈ N, therefore implies

if n ≥ N, then s −  < sN ≤ sn ≤, which shows that {sn} 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|, . . . , |sK−1|, 1 + |s|} =⇒ |sn| ≤ L ∀ n ∈ N.

This proves that {sn} 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 {In} such that the right endpoint of In is an upper bound of S, In∩ S 6= ∅ ∀ n ≥ 1.

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and

I1 ⊃ I2 ⊃ · · · ⊃ In⊃ · · · ; lim

n→∞|In| = lim

n→∞

M 2n−2 = 0,

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

n→∞lim xn = x.

Since lim

n→∞|In| = lim

n→∞

M

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

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

n→∞lim xn = x.

Upper and Lower Limits

Definition Let {sn} be a sequence of real numbers.

(a) lim

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

(b) lim

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

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

E =n

x ∈ (−∞, ∞) ∪ {±∞} | ∃ {snk} ⊂ {sn} such that lim

k→∞snk = xo . Let

lim sup

n→∞

sn = s = sup E, lim inf

n→∞ sn = s = inf E.

The numbers s, s are called the upper and lower limits of {sn}; 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 {sn}. Then s = sup E and s = inf E have the following properties:

(a) s, s ∈ E, i.e. there exist subsequences {snk}, {smj} of {sn} such that lim

k→∞snk = s, lim

j→∞smj = s. Proof

If s = ∞, then E is not bounded above; hence {sn} is not bounded above, and there is a subsequence {snk} such that lim

k→∞snk = ∞.

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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 sn > 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−  < sn< s+ .

s+  s − 

Proof For each  > 0, if there are infinitely many sn’s lying in (−∞, s− ] ∪ [s+ , ∞), then there exists a subsequence {snk} 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 xn’s lying in (−∞, s− ] ∪ [s+ , ∞).

Remarks

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

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

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

(b) If s, s ∈ E, and {snk}, {smj} are subsequences of {sn} such that

k→∞lim snk = s, lim

j→∞smj = s =⇒ lim inf

k→∞ snk = s = lim sup

k→∞

snk, lim inf

j→∞ smj = s = lim sup

j→∞

smj. For each  > 0, since there only exist finite number of n ∈ N such that sn > s +  or sn < s− , hence we have

s−  ≤ inf{sm | m ≥ n} ≤ sup{sm | m ≥ n} ≤ s+  for sufficiently large n

=⇒ s−  ≤ lim

n→∞inf{sm | m ≥ n} ≤ lim

n→∞sup{sm | m ≥ n} ≤ s+  ∀  > 0

=⇒ lim

→0+(s− ) ≤ lim

n→∞inf{sm | m ≥ n} ≤ lim

n→∞sup{sm | m ≥ n} ≤ lim

→0+(s+ )

=⇒ s ≤ lim

n→∞inf{sm | m ≥ n} ≤ lim

n→∞sup{sm | m ≥ n} ≤ s.

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

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

=⇒ inf{sn| n ≥ ml} ≤ lim

j→∞smj = s and s = lim

k→∞snk ≤ sup{sm | m ≥ nl} ∀ l ∈ N

=⇒ inf{sn| n ≥ ml} ≤ s ≤ s ≤ sup{sm | m ≥ nl} ∀ l ∈ N

=⇒ lim

l→∞inf{sn | n ≥ ml} ≤ s ≤ s ≤ lim

l→∞sup{sm | m ≥ nl}

=⇒ lim

n→∞inf{sm | m ≥ n} ≤ s ≤ s ≤ lim

n→∞sup{sm | m ≥ n}.

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Hence

lim sup

n→∞

sn= s = lim

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

n→∞ sn= s = lim

n→∞inf{sm | m ≥ n} = sup{inf{sm| 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{sm | m ≥ n} − s|, |s− inf{sm | m ≥ n}|} < min{x − s, s− y} ∀ n ≥ N

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

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

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

Examples

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

lim sup

n→∞

sn= ∞, lim inf

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

(b) Let sn= (−1)n

1 + (1/n). Then

lim sup

n→∞

sn = 1, lim inf

n→∞ sn = −1.

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

n→∞lim sn = s ⇐⇒ lim sup

n→∞

sn = lim inf

n→∞ sn = s.

Theorem If sn ≤ tn for n ≥ N, where N is fixed, then lim inf

n→∞ sn≤ lim inf

n→∞ tn, lim sup

n→∞

sn ≤ lim sup

n→∞

tn.

Some Special Sequences

Theorem

(a) If p > 0, then lim

n→∞

1 np = 0.

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

1/p ⇐⇒ np > 1

 ⇐⇒  > 1 np = | 1

np − 0|.

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(b) p > 0, then lim

n→∞

n

p = 1.

Proof If p > 1, by letting xn = √n

p − 1 > 0, then 1 + n xn≤ (1 + xn)n= p =⇒ xn ≤ p − 1

n =⇒ lim

n→∞xn = 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 xn = √n

n − 1. Then xn≥ 0 for each n ∈ N and n = (1 + xn)n≥ n(n − 1)

2 x2n =⇒ 0 ≤ xn

r 2

n − 1 for n ≥ 2 =⇒ lim

n→∞xn = 0.

(d) If p > 0 and α ∈ R, then lim

n→∞

nα

(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



pk = n(n − 1) · · · (n − k + 1)

k! pk> nkpk

2kk! =⇒ 0 < nα

(1 + p)n < 2kk!

pk nα−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 {an} be a real sequence. Then:

j

X

n=i

an = ai+ ai+1+ · · · + aj, when i < j ∈ N.

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Definition The nth partial sum of {ak} is sn=

n

X

k=1

ak.

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

Definition If sn → s we write

X

n=1

an = lim

n→∞

n

X

k=1

ak = 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 sn=

n

X

k=1

1

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

= sm+1+ sm+2+ · · · + sn

= 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 s2n − sn ≥ 2n − n

2n = 1

2. Therefore this sequence is not Cauchy. Hence {sn} 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

ak

< .

Corollary (Divergence Test) If X

an converges, then lim

n→∞an = 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 an ≥ 0, then X

an converges if and only if partial sums {sn} form a bounded sequence.

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

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

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Comparison Test 1. If X

cn converges and |an| ≤ cn for almost all n, then X

an converges.

2. If X

dn diverges to +∞ and an ≥ dn for almost all n, then X

an diverges as well.

Proof

1. Let  > 0. Since X

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

m

X

k=n

ck

m

X

k=n

ck

< .

Thus

m

X

k=n

ak

m

X

k=n

|ak| ≤

m

X

k=n

ck.

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

an 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 sn = 1

1 − x if |x| < 1

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

Example

X

n=0

1

n! converges. Notice

sn= 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.

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Definition e =

X

n=0

1 n!. Theorem e = lim

n→∞(1 + 1 n)n. Proof

Let sn=

n

X

k=0

1

k!, tn = 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 tn≤ sn ≤ e for all n ∈ N, so that

lim sup

n→∞

tn≤ e.

Next, if n ≥ m, then

tn≥ 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→∞ tn ≥ 1 + 1 + 1

2! + · · · + 1

m! =⇒ lim inf

n→∞ tn ≥ sm ∀ m ∈ N.

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

n→∞ tn ≤ lim sup

n→∞

tn≤ e =⇒ e = lim

n→∞tn = lim

n→∞(1 + 1 n)n.

Remark

e − sn = 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 − sn)

| {z }

N

< 1 n < 1

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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

an con-

verges if and only if

X

k=0

2ka2k = a1+ 2a2+ 4a4 + 8a8+ . . . converges.

Proof Compare sn = a1 + · · · + an and tk = a1 + 2a2 + · · · + 2ka2k. Consider the following grouping of the finite sum:

sn = (a1) + (a2+ a3) + (a4 + · · · + a7) + · · · + an

tk = (a1) + (a2+ a2) + (a4 + · · · + a4) + · · · + (a2k + · · · + a2k)

If tk converges, since sn < tk ∀ n < 2k and ∀ k ∈ N,

=⇒ sn≤ lim

k→∞tk ∀ n ∈ N

=⇒ sn converges since sn is monotonically increasing and bounded above.

On the other hand,

if sn converges, since 2sn> tk− a1 ∀ n ≥ 3 and ∀ 2k < n,

=⇒ 2 lim

n→∞sn ≥ tk− a1 ∀ k ∈ N

=⇒ tk converges since tk− a1 is monotonically increasing and bounded above.

This proves that both series diverge or converge simultaneously.

Theorem Consider P 1

np. 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

2k 1

(2k)p =X

k

2(1−p)k,

which is geometric. Thus it converges if and only if 21−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 an converges;

(b) if α > 1, then P an diverges;

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

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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 {nk} such that

nk

q

|ank| → α > 1.

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

n→∞lim an6= 0 =⇒ X

an 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

an < 1, (b) diverges if

an+1

an

≥ 1 for n ≥ n0, where n0 is some fixed integer.

Proof If lim sup

an+1 an

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

an+1

an

< β < 1 for n ≥ N.

This implies that

|aN +k| < β|aN +k−1| < β2|aN +k−2| < · · · < βk|an| for each k ∈ N.

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

If |an+1| ≥ |an| for n ≥ n0, then

n→∞lim an6= 0 =⇒ X

an diverges.

Definition A power series is of the form

X

k=0

cnzn where cn ∈ 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.

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Theorem P an converges absolutely implies P an converges.

Proof |

m

X

k=n

ak| ≤

m

X

k=n

|ak| < , by the Cauchy Criterion since P |ak| converges.

Example If a series converges, it does not necessarily converge absolutely. Consider the series 1 − 12+1314 + . . . , 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 an converges but not absolutely, then we can form a rear- rangement that has any limsup and liminf you’d like.

Example Rearrange 1 − 12 + 1314 + . . . 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.

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