Part II
Basic Analysis
6 Convergence of Sequences and Series
A real-valued sequence (snn: n EN) is a functionnsn: N JR, and so extends
to a hypersequence s : *N *JR by the construction of Section 3.n13. Hence
the term sn becomes defined for unlimited hypernaturals n E *N00 (a fact
that was already used in Theorem 5.8.1), and in this case we say that Sn
is an extended term of the sequence. The collection
{sn
: n E *Noo}
of extended terms is the extended tail of s.
6.1 Convergence
In real analysis, (sn : n EN) converges to the limit L E 1R when each open
interval (Lo-c, L +c) around L in 1R contains some standard tail of the
sequence, i.e., contains all the terms
from some point on (with this point depending on c). Formally, this is
expressed by the statement
(Vc E JR+) (:Jmc EN) (Vn EN) (n >me isn-Ll me lsn
-Ll me lsn-Ll me because N is unlimited and me is limited, and so this
last sentence implies !sN -Ll < c as desired.
For the converse, suppose sn L for all unlimited n. We have to show
that any given interval (L -c, L + c) in lR contains some standard tail of
the sequence. The essence of the argument is to invoke the fact that the
extended tail is infinitely close to L, hence contained in *(L
-c, L +c), and
then apply transfer.
To spell this out, fix an unlimited N E *N00• Then for any n E *N,
if n > N, it follows the!t n is also unlimited, so sn L and therefore
lsn
-Ll N lsn
-Ll z lsn
-Ll z lsn
-Ll oo tn = M in lR, then
(1) limn---.oo(Sn + tn) = L + M,
(2) limn__.00n(csn) = cL, for any c E lR,
(3) limn__.00(Sntn) = LM,
(4) limn__.00(snftn) = LjM, if M =/= 0.
Proof Use Exercise 5.7(1). D
6.4 Boundedness and Divergence
Theorem 6.4.1 A real-valued sequence (sn) is bounded in lR if and only
if its extended terms are all limited.
Proof To say that (sn : n EN) is bounded in lR means that it is contained
within some real interval [-b, b], or equivalently that its absolute values lsnl
have some real upper bound b:
('v'n E N) lsnI < b.
Then by universal transfer the extended sequence is contained in *[-b, b],
i.e., lsnl < b for all n E *N; hence sn is limited in general.
For the converse, if Sn is limited for all unlimited n E *N00
, then it is
limited for all n E *N. Hence if r E *JR;t, is any positive unlimited hyperreal,
we observe that the entire extended sequence lies in the interval {x E *JR
:
-r < x < r} and apply transfer. More formally, we have lsnI < r for all
n E *N, so the sentence
(3y E *IR) (\In E *N) lsnl < Y
·
isntrue. But then by existential transfer it follows that there is some real
number that is an upper bound to lsnl for all n EN. D
This proof can be refined to show the following:
• the real-valued sequence (sn) is bounded above in IR, i.e., there is a
real upper bound to {sn : n E N}, if and only if it has no positive
unlimited extended terms;
• (sn) is bounded below in JR, i.e., there is a real lower bound to {sn :
n EN}
, if and only if it has no negative unlimited terms.
We say that (sn) diverges to infinity if for each real r there is some n EN
such that all terms of the standard tail sn, Sn+b sn+2, .n.. are greater than
r. Correspondingly, (sn) diverges to minus infinity if for each real r there
is some n E N such that sm < r for all m :2:: n.
6.5 Cauchy Sequences 65
Theorem 6.4.2 A real-valued sequence
(1) diverges to infinity if and only if all of its extended terms are positive
unlimited; and
(2) diverges to minus infinity if and only if all of its extended terms are
negative unlimited.
Proof Exercise. 0
6.5 Cauchy Sequences
The standard definition of a Cauchy sequence is one that satisfies
lim lsn-sml = 0,
m,n->oo
meaning that the terms get arbitrarily close to each other as we move along
the sequence. Formally this is rendered by the sentence
(Vc E JR+) (::Jj EN) ('Vm, n EN) (m, n 2': jo---+ I
sm-snl m 1\ isno-Ll m, and hence
n is unlimited, and
lsn Ll < e 0.
-
Thus sn is an extended term infinitely close to L. (Indeed, the argument
shows that any interval of infinitesimal width around L contains terms
arbitrarily far along the extended tail.)
Conversely, suppose there is an unlimited N with SN L. To prove (i),
take any positive c: E 1R and mEN. Then N > m and isNo-Ll <
e. This
shows that
(3n E *N) (n > m
l\ lsno-Ll < c:).
Thus by existential transfer, lsno-Ll < e for some n EN, with n > m. D
This characterisation shows that a shadow of an extended term is a cluster
point of a real sequence, and indeed that the cluster points are precisely
the shadows of those extended terms that have them, i.e., are limited. But
if the sequence is bounded, then all of its extended terms are limited and
so have shadows that must be cluster points. In particular, this gives a very
direct proof of a famous result:
Theorem 6.6.2 (Bolzano-Weierstrass) Every bounded sequence of real
numbers has a cluster point in JR. D
6.7 Exercises on Limits and Cluster Points
(1) Let (sn) and (tn) be real-valued sequences with limits L, M respectively.
Show that if Sn :::; tn for n E N, then L :::; M.
6.8 Limits Superior and Inferior 67
(2) If rn Sn tn in 1R for all EN, and limn--+oo rn = limn--+oo tn, show
that (sn) converges to this same limit.
(3) If a sequence converges in JR, show that it has exactly one cluster
point.
(4) Suppose that a real-valued sequence has a single cluster point. If the
sequence is bounded, must it be convergent? What if it is unbounded?
6.8 Limits Superior and Inferior
Let s = (sn : n E N) be a bounded real-valued sequence. Even if the
sequence does not converge, its behaviour can be analysed into patterns
of regularity: the Bolzano-Weierstrass theorem guarantees that it has a
cluster point, and it may have many such cluster points with subsequences
converging to each of them. If Cs is the set of all cluster points of s, then
by the characterisation of Theorem 6.6.1 it can also be described as the set
of all shadows of the extended tail:
Cs = {sh(sn) : n is unlimited}.
Now, any real upper bound of the original sequence is an upper bound of
C8n, for if sn b for all n E N (with b E JR), then for an unlimited n we
get sh(sn) Sn b and hence sh(sn) b, since both are real. Similarly,
any real lower bound of s is a lower bound of C 8• Since s is a bounded
sequence, it follows that the set C8 is bounded above and below in JR, and
so has a real least upper bound, known as the limit superior of s, and a
real greatest lower bound, known as the limit inferior. The notations
lim sup Sn and lim inf Sn
n--+oo n--+oo
are used for these two numbers.
Writing "sup" for the least upper bound (supremum) and "inf" for the
greatest lower bound (infimum), we have
limnsupnsn sup{sh(sn)n: n E *Noo},
n--+oo
limninf Sn -inf{sh(sn) : n E *N00}. n---+oo
The symbols lim and lim are also used for limnsup and limninf.
Exercise 6.8.1
Prove, by nonstandard reasoning, that both the limit superior and the limit
inferior are cluster points of the sequence s.
68 6. Convergence of Sequences and Series
This exercise shows that lim s and lim s both belong to Csn, and hence are
the maximum and minimum elements of C8 respectively.
Theorem 6.8.2 A real number L is equal to lim s if and only if
(1) Sn < L or Sn L for all unlimited n; and
(2) Sn L for at least one unlimited n.
Proof The condition "sn < Lor Sn L" holds iff sh(sn) :S L. Thus (1) is
equivalent to the assertion that Lis an upper bound of C8n• But (2) asserts
that Lis a cluster point, so (1) and (2) together assert precisely that L is
the maximum element of C8, i.e., that L = limns. 0
Formulation of the nonstandard characterisation of lim s corresponding to
Exercise 6.8.1 is left as a further exercise.
Theorem 6.8.3 A bounded real-valued sequence s converges to L E 1R if
and only if
limsupnsn = liminf Sn = L.
n-+oo
n-+oo
Proof. Since lim s and lim s are the maximum and minimum elements
of C8n, requiring that they both be equal to L amounts to requiring that
Cs = { L }. But that just means that the shadow of every extended term
is equal to L, which is equivalent to having s converge to L by Theorem
6.1.1. D
Theorem 6.8.4 If s is a bounded real-valued sequence with limit superior
(1) some standard tail of s has all its terms smaller than limn+ c, i.e.,
Sn < lim + c for all but finitely many n E N;
Proof.
then for any positive real c:
(2) limn-c < Sn for infinitely many n EN.
(1) If mE *N is unlimited, then sh(sm) :::;: lim, so
Sm sh(sm) oo Bn. Then L is the greatest lower bound of the
sequencenS, and by (ii), lim is a lower bound for this sequence, so limn::; L.
But if lim < L, we can choose a positive real c with lim + c < L, and
then by Theorem 6.8.4( 1) there is some n E N such that the standard tail
sn, Sn+l, Bn+2, ... is bounded above by lim+ c. This implies that the least
upper bound Bn of this tail is no bigger than limn+ c. However, that gives
Bn ::; lim + c < L,
which contradicts the fact that Lis a lower bound of S, and so L::; Bn·
We are left with the conclusion that lim = L, as desired. 0
The notion of limit superior can be defined for any real-valued sequence
s = (sn), bounded or not, by a consideration of cases in the following way.
(1) If s is not bounded above, put limsupn->oo Sn = +oo. In this case s
has at least one positive unlimited extended term (Section 6.4).
(2) If s is bounded above, hence has no positive unlimited extended
terms, then there are two subcases:
(i) s diverges to minus infinity. Then put limnsupn->oo Sn = -oo. In
this case all extended terms are negative unlimited (Theorem
6.4.2), so there are no limited extended terms and therefore no
cluster points.
(ii) s does not diverge to minus infinity. Then there is at least one extended
term that is not negative unlimited, and hence is limited
because there are no positive unlimited terms (as s is bounded
above). The shadow of this term is a cluster point of s. Thus
the set C8 of cluster points is nonempty and bounded above (by
any upper bound for s). Then we define limsupn->oo Sn to be
the least upper bound of Cs as previously.
6.9 Exercises on lim sup and lim inf
(1) Formulate the definition of the limit inferior of an arbitrary realvalued
sequence.
(2) Formulate and prove theorems about the limit inferior of a bounded
sequence that correspond to Theorems 6.8.4 and 6.8.5.
(3) If sis a bounded sequence, show that for each c E JR+ there exists an
n E N such that the standard tail .Sn, Sn+b Bn+2, ... is contained in
the interval ( lim c, lim + c ).
-
6.11 Exercises on Convergence of Series
6.10 Series
A real infinite series 2::: ai is convergent iff the sequence s = (sn : n E N)
of partial sums
Sn = a1 + ... +nan
is convergent. We write 2::: ai for sn, and 2:::: ai for Sn-sm-1 when n m.
Extending s to a hypersequence (sn : n E *N), we get that Sn and sm-l
are defined for all hyperintegers n, m, so the expressions 2::: ai and 2:::: ai
become meaningful for all n, mE *N, and may be thought of as hyperfinite
sums when n is unlimited.
Applying our results on convergence of sequences to the sequence of
partial sums, we have:
• 2::: ai = L in 1R iff 2::: ai L for all unlimited n.
• 2::: ai converges in 1R iff 2:::: ai 0 for all unlimited m, n with
m
n.
The second of these is given by the Cauchy convergence criterion (Theorem
6.5.2), since 2:::: ai 0 iff Sn Sm-1 for unlimited m, n. Taking the
case m = n here, we get that if the series 2::: ai converges, then an 0
whenever n is unlimited. This shows, by Theorem 6.1.1, that
• if 2::: ai converges, then limi-+oo ai = 0.
0 bserve that for a convergent real series we have
L ai = sh (2::: ai)
for any unlimited n.
A series 2::: ai always converges if it is absolutely convergent, which
means that the series 2::: laiI of absolute values converges. The standard
proof of this uses the comparison test, which is itself illuminated by nonstandard
ideas (see Exercise ( 4) below).
6.11 Exercises on Convergence of Series
(1) Give an example of a series that diverges but has an infinitesimal for
all unlimited n.
(2) Give nonstandard proofs of the usual rules for arithmetically combining
convergent series:
2::: ai + 2::: bi L(ai + bi),
2::: ai 2::: bi -L(ai-bi),
2::: cai c (2::: ai) .
-
72 6. Convergence of Sequences and Series
(3) Suppose that ai 0 for all i E N. Prove that 2: ai converges iff
2: ai is limited for all unlimited n, and that this holds iff 2: ai is
limited for some unlimited n.
(4) (Comparison Test) Let 2: ai and 2: bi be two real series of nonnegative
terms, with 2: bi convergent. If ai ::; bi for all i E N, use
result (3) to show that 2: ai is convergent.
(5) Show that the comparison test holds under the weaker assumption
that an ::; bn for all unlimited n (hint: use the Cauchy convergence
criterion). Show that this weaker assumption is equivalent to requiring
that there be some limited kEN with an ::; bn for all n k.
(6) Let 2: ai and 2: bi be two series of positive terms such that the
sequence (adbi : i E N) is convergent in JR. Show that for unlimited
m and n, 2: ai is infinitesimal if and only if 2: bi is infinitesimal.
Deduce that either both series converge, or both diverge.
(7) Let c E R Recall the identity
1-cn+l
1 + c+ c2
1-c
(a) Considering the case of unlimited n, show that the series 2: ci
converges if lei < 1.
(b) Show that 2: ci is infinitesimal when m and n are unlimited,
either by applying result (a) or by making further use of the
above identity.
(8) (Ratio Test: Convergence) Suppose that
a
lai+l
l
lisup < 1
-+oo
l
l
in lR (i.e., the limit superior of the sequence of ratios is a real number
smaller than 1). Prove that the series 2: ai is absolutely convergent,
by the following reasoning.
(a) Show that there exists a positive real c < 1 with lan+ll < c !ani
for all unlimited n.
(b) Hence show that there is some limited k E N such that lan+ll <
c ianI for all n k.
(c) Deduce from (b) that in general,
iak+ni <
en laki
'
and hence
6.11 Exercises on Convergence of Series
(d) Use the last inequality and result 7(b) to conclude that E ai
converges absolutely.
(9) (Ratio Test: Divergence) Suppose that
l. . lai+II
Imnmf > 1
t-+oo
ai
in ffi. (i.e., the lim inf is a real number greater than 1). Prove that the
series E ai diverges, as follows.
(a) Show that lan+ll > lanl for all unlimited n. Hence prove that
there is some limited k E N such that lanl > lakl > 0 for all
n > k.
(b) Deduce, by considering unlimited n, that E ai cannot converge.
(10) Apply the ratio test with ai = (x
i/i!) to show that for any real
number x, the hyperreal xn jn! is infinitesimal when n is unlimited.
(11) (Leibniz's Alternating Series Test) Suppose (ai : i E N) is a real
sequence that is nonincreasing (i.e., ai 2: ai+1) and converges to 0.
Prove that the alternating series
converges by showing that in general
and then considering the case of unlimited m.