Subsequences Some of the terms of the sequence {pn} in order.
Definition Let {pn} be a sequence in X. 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 from last class, almost all terms of the subsequence are contained in any neigh- borhood of the limit.
Remarks
(a) Note If pn → p, then pnk → p for any subsequence {pnk} of {pn}.
(b) Question Must every sequence contain a convergent subsequence?
No. Consider the sequence {1, 2, 3, . . . }. There is no convergent subsequence of this se- quence.
Theorem In a compact metric space X, every sequence contains a subsequence converging to a point in X.
Corollary Every bounded sequence in Rkcontains a convergent subsequence. (Bolzano-Weirstrauss).
Proof Let R = range({pn}). If R is finite, then some p ∈ R is hit infinitely many times by the sequence by the pigeon-hole principle. Say {pn1, pn2, . . . } are those terms. Then this {pnk} converges to p. If R is infinite, then since X is compact, R has a limit point. Use property E to get a subsequence converging to p.
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) < .
Theorem If {pn} converges, then it is Cauchy.
Proof Say 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).
Definition A metric space X is complete if every Cauchy sequence converges to a point in X.
Theorem 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.
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 the upper and lower limits of {sn}, denoted by lim sup
n→∞
sn and lim inf
n→∞ sn respectively, be defined by
lim sup
n→∞
sn = sup E, lim inf
n→∞ sn = inf E.
By the Theorem 3.17, the numbers s∗ = sup E and s∗ = inf E satisfy the following properties:
(a) s∗, s∗ ∈ E, i.e. there exist subsequences {snk}, {smj} of {sn} such that
k→∞lim snk = s∗, lim
j→∞smj = s∗. (b) For each > 0, there exists an N ∈ N such that
if n ≥ N then s∗− < sn< s∗+ .
Remark Note that
(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}.
(b) lim
n→∞sup{sm | m ≥ n} and lim
n→∞inf{sm | m ≥ n} are points in E, i.e. there exist subse- quences {snk} and {smj} of {sn} such that
lim
k→∞snk = lim
n→∞sup{sm | m ≥ n} and lim
j→∞smj = lim
n→∞inf{sm | m ≥ n}.
Thus, lim inf
n→∞ sn = inf E ≤ lim
n→∞inf{sm | m ≥ n} ≤ lim
n→∞sup{sm | m ≥ n} ≤ sup E = lim sup
n→∞
sn. Since {nk | n1 < n2 < . . . → ∞} ⊆ {m ∈ N | m ≥ n1}, a subsequence {snk} of {sn} is a subsequence of {sm | m ≥ n1} = {sn1, sn1+1, sn1+2, . . .},
=⇒ inf{sm | m ≥ n1} ≤ snk ≤ sup{sm | m ≥ n1} for all k ∈ N,
=⇒ inf{sm | m ≥ n1} ≤ lim inf
k→∞ snk ≤ lim sup
k→∞
snk ≤ sup{sm | m ≥ n1}.
Hence,
n→∞lim inf{sm | m ≥ n} ≤ lim inf
n→∞ sn≤ lim sup
n→∞
sn ≤ lim
n→∞sup{sm | m ≥ n}, and
lim sup
n→∞
sn = lim
n→∞sup{sm | m ≥ n}, lim inf
n→∞ sn = lim
n→∞inf{sm | m ≥ n}.
Theorem 3.11(2) Let X be a compact metric space and let {pn} be a sequence of points in X.
If {pn} is Cauchy then {pn} is convergent.
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}
denote the minimum distance of the m(m − 1)/2 pair of points. Let δ > > 0 be given.
Since {pn} is Cauchy, there exists N ∈ N such that
if n, m ≥ N then d(pn, pm) < < δ.
Suppose ∃ n, m ≥ N with pn6= pm, we get δ ≤ d(pn, pm) < δ which is a contradiction.
Hence, we have
∀ n, m ≥ N, pn= pm,
i.e. {pn| n ≥ N }, hence {pn| n ∈ N}, is a convergent sequence that converges to pN. Case 2: S contains infinitely many points.
Since X is compact, X contains all the limit points of S, i.e. S0 ⊂ X. Let p be a limit point of S.
For each k ∈ N, there exists pnk ∈ N1/k(p) ∩ S \ {p}. Then {pnk} is a subsequence of {pn} converging to p, i.e. for each > 0, there exists N1 ∈ N such that if k ≥ N1 then d(pnk, p) < .
Also, since {pn} is a Cauchy sequence, there exists 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 for all n ≥ N, we have proved that lim
n→∞pn= p.
Example R is complete but Q is not.
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)).
Example If X = Q, then X? = R. In particular, X? is isometrically isomorphic to R. In other there is a distance preserving bijection.
Remark This is the other construction of R. But Dedekind cuts are more hardcore.
Definition A sequence {sn} in R is monotonically increasing if sn ≤ sn+1 for all n. Similarly {sn} is monotonically decreasing if sn ≥ sn+1.
Theorem Let {sn} is monotonic (i.e., either). Then {sn} converges in R if and only if it is bounded.
Proof (⇐) Suppose sn ≤ sn+1without loss of generality. Let s = sup(range({sn})). Then sn ≤ s for all n.
Let > 0. Then s − < s. Thus there exists N ∈ N such that s − < sN ≤ s since s is a least-upper-bound by construction. Furthermore, since sn+1 ≥ sn for all n. Thus if n > N implies
s − < sN ≤ sn≤ s.
So d(sn, s) < . Therefore sn → s.
(⇒) Already have shown that convergence sequences are bounded.
Definition Let {sn} be a sequence in R. We define the upper limit of {sn} in R ∪ {±∞} (i.e.
the completion of R), is
n→∞lim sup{sk|k > n} = lim sup sn. Similarly the lower limit of sn is:
n→∞lim inf{sk|k > n} = lim inf sn.
Theorems
(a) (Book definition) Let {sn} be real. Let E be the set of all sub sequential limits of {sn}. In other words:
E = {x ∈ R ∪ {±∞}|snk → x for some subsequence {snk}}.
Let s? = sup E, s? = inf E. Then s? = lim sup sn and s? = lim inf sn. (b) sn → s ⇐⇒ lim sup sn= lim inf sn= s.
(c) Let {sn} be real. Then
(a) lim sup sn∈ E, using the definition of E above.
(b) If x > lim sup sn, then ∃N ∈ N such that n > N implies sn < x. Moreover, lim sup sn is the only number with this property. It is analogous for lim inf sn.
Proof (Idea) Let snk → t ∈ E. Then lim inf snk = lim sup snk = t. But {snk} ⊆ {sn}.
Notice:
lim inf sn≤ lim inf snk = t = lim sup snk ≤ lim sup sn. Thus
lim inf sn ≤ inf E ≤ sup E ≤ lim sup sn.
But lim inf sn∈ E, lim inf sn∈ E. So lim inf sn = inf E, lim sup sn = sup E.
Example Special Sequences. Let p > 0.
(a) lim
n→∞
1 np = 0.
(b) lim
n→∞pn1 = 1.
(c) lim
n→∞nn1 = 1.
(d) lim
n→∞
na
(1 + p)n = 0 for all a ∈ R.
(e) If |x| < 1, then lim
n→∞xn = 0.
Series
Question What does this mean?
1 1 1 1 π2
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.
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 {sn} is monotonic. Thus bounded implies and is implied by convergence.
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.
X n
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
< 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!. Remark
e − sn = 1
(n + 1)! + 1
(n + 2)! + . . .
= 1
(n + 1)!(1 + 1
(n + 1)! + 1
(n + 2)! + . . . )
= 1 n!n
Theorem e 6∈ Q.
Proof Suppose e = m
n for m, n > 0. Then:
0 ≤ n!(e − sn)
| {z }
∈Z
< 1 n < 1
Remark e = lim
n→∞(1 + 1 n)n.
Recall A series converges if its partial sums converge.
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, thenP anconverges if and only if P 2ka2k = a1+ 2a2+ 4a4+ 8a4 + . . . 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 n < 2k, then sn < tk. If n > 2k, then 2sn> tk. Compare then:
2a1+ 2a2+ 2(a3+ a4) + . . . a1+ (a2+ a2) + 4a4+ . . . So both series diverge or converge together.
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.
Root Test Given P an Let α = lim
n→∞
p|an n|. If α < 1, then P an converges. If α > 1, then P an diverges. If α = 1, then the test is inconclusive. However this is unsatisfactory. So let α = lim sup
n→∞
p|an n|, which always exists (though it may be infinity).
Proof (By comparison with the geometric series) Suppose α < 1. Remark that if a sequence passes a limsup, then it passes it for only finitely many terms. Choose α < β < 1. By the definition of limsup there exists N such that n > N implies p|an n| < β. So |an| < βn. ButP bn converges. Therefore, the sum of the an converges as well.
If α > 1, then there exists a subsequence of p|a4 n|, say akp|ank → α > 1. So akp|ank > 1 for k large enough. This implies |ank| > 1 and therefore the terms do not go to zero and thus the sequence does not converge.
If α = 1, notice P 1
n diverges but P 1
n2 converges but in both cases α = 1 and therefore this case is inconclusive.
Ratio Test P an converges of lim sup
an+1 an
< 1, diverges if
an+1 an
≥ 1 for n large enough.
Proof If the ratio is always bigger than 1 so the series doesn’t converge. If the ratio is smaller than 1, then we know
an+1
a
< β < 1 for some large N . So aN +k < βaN +k−1 < β2aN +k−2 <
· · · < βkan. So compare:
∞
X
k=0
aN +k an
∞
X
k=0
βk.
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.
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+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 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 + 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.
