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Whatwe know about the sequence in = infk≥n xk is that it is nondecreasing and thatlimn→∞ in = i ∈ R. For the numbers n ∈ N, using the definition of the greatestlower bound, we choose by induction numbers kn ∈ N such that kn < kn+1 andikn ≤ xkn < ikn + n1 . (Taking i1 we find k1 ; taking ik1 +1 we find k2 , etc.) Sincelimn→∞ in = limn→∞ (in + n1 ) = i, we can assert, by properties of limits, thatlimn→∞ xkn = i. We have thus proved that i is a partial limit of the sequence {xk }.It is the smallest partial limit since for every ε > 0 there exists n ∈ N such thati − ε < in , that is i − ε < in = infk≥n xk ≤ xk for any k ≥ n.The inequality i − ε < xk for k > n means that no partial limit of the sequencecan be less than i − ε.
But ε > 0 is arbitrary, and hence no partial limit can be lessthan i.The proof for the superior limit is of course analogous.We now remark that if a sequence is not bounded below, then one can select asubsequence of it tending to −∞. But in this case we also have limk→∞ xk = −∞,and we can make the convention that the inferior limit is once again the smallestpartial limit.
The superior limit may be finite; if so, by what has been proved itwe are assuming the natural relations −∞ < x < +∞ between the symbols −∞, +∞ andnumbers x ∈ R.4 Here943Limitsmust be the largest partial limit. But it may also be infinite. If limk→∞ xk = +∞,then the sequence is also unbounded from above, and one can select a subsequencetending to +∞. Finally, if limk→∞ xk = −∞, which is also possible, this meansthat supk≥n xk = sn → −∞, that is, the sequence {xn } itself tends to −∞, sincesn ≥ xn .
Similarly, if limk→∞ xk = +∞, then xk → +∞.Taking account of what has just been said we deduce the following proposition.Proposition 1 For any sequence, the inferior limit is the smallest of its partiallimits and the superior limit is the largest of its partial limits.Corollary 3 A sequence has a limit or tends to negative or positive infinity if andonly if its inferior and superior limits are the same.Proof The cases when limk→∞ xk = limk→∞ xk = +∞ or limk→∞ xk =limk→∞ xk = −∞ have been investigated above, and so we may assume thatlimk→∞ xk = limk→∞ xk = A ∈ R. Since in = infk≥n xk ≤ xn ≤ supk≥n xk = snand by hypothesis limn→∞ in = limn→∞ sn = A, we also have limn→∞ xn = A byproperties of limits.Corollary 4 A sequence converges if and only if every subsequence of it converges.Proof The inferior and superior limits of a subsequence lie between those of thesequence itself.
If the sequence converges, its inferior and superior limits are thesame, and so those of the subsequence must also be the same, proving that thesubsequence converges. Moreover, the limit of the subsequence must be the sameas that of the sequence itself.The converse assertion is obvious, since the subsequence can be chosen as thesequence itself.Corollary 5 The Bolzano–Weierstrass Lemma in its restricted and wider formulations follows from Propositions 1 and 1 respectively.Proof Indeed, if the sequence {xk } is bounded, then the points i = limk→∞ xk ands = limk→∞ xk are finite and, by what has been proved, are partial limits of thesequence. Only when i = s does the sequence have a unique limit point. Wheni < s there are at least two.If the sequence is unbounded on one side or the other, there exists a subsequencetending to the corresponding infinity.Concluding Remarks We have carried out all three points of the program outlinedat the beginning of this section (and even gone beyond it in some ways).
We havegiven a precise definition of the limit of a sequence, proved that the limit is unique,explained the connection between the limit operation and the structure of the set ofreal numbers, and obtained a criterion for convergence of a sequence.We now study a special type of sequence that is frequently encountered and veryuseful – a series.3.1 The Limit of a Sequence953.1.4 Elementary Facts About Seriesa. The Sum of a Series and the Cauchy Criterion for Convergence of a SeriesLet {an } be a sequence of real numbers.We recall that the sum ap + ap+1 + · · · +aq , (p ≤ q) is denoted by the symbol qn=p an . We now wish to give a precisemeaning to the expression a1 + a2 + · · · + an + · · · , which expresses the sum of allthe terms of the sequence {an }.Definition16 The expression a1 + a2 + · · · + an + · · · is denoted by the symbol∞aandusually called a series or an infinite series (in order to emphasize itsnn=1difference from the sum of a finite number of terms).Definition 17 The elements of the sequence {an }, when regarded as elements of theseries, are called the terms of the series.
The element an is called the nth term.Definition 18 The sum sn = nk=1 ak is called the partial sum of the series, or,when one wishes to exhibit its index, the nth partial sum of the series.5Definition 19 If the sequence {sn } of partial sums of a series converges, we say theseries is convergent. If the sequence {sn } does not have a limit, we say the series isdivergent.Definition 20 The limit limn→∞ sn = s of the sequence of partial sums of the series, if it exists, is called the sum of the series.It is in this sense that we shall henceforth understand the expression∞an = s.n=1Since convergence of a series is equivalent to convergence of its sequence ofpartial sums {sn }, applying the Cauchy convergence criterion to the sequence {sn }yields the following theorem.Theorem 6 (The Cauchy convergence criterion for a series) The series a1 + · · · +an + · · · converges if and only if for every ε > 0 there exists N ∈ N such that theinequalities m ≥ n > N imply |an + · · · + am | < ε.Corollary 6 If only a finite number of terms of a series are changed, the resultingnew series will converge if the original series did and diverge if it diverged.we are actuallydefining a series to be an ordered pair ({an }, {sn }) of sequences connectedby the relation (sn = nk=1 ak ) for all n ∈ N.5 Thus963LimitsProof For the proof it suffices to assume that the number N in the Cauchy convergence criterion is larger than the largest index among the terms that were altered.
Corollary 7 A necessary condition for convergence of the series a1 + · · · + an + · · ·is that the terms tend to zero as n → ∞, that is, it is necessary that limn→∞ an = 0.Proof It suffices to set m = n in the Cauchy convergence criterion and use the definition of the limit of a sequence.Here is another proof: an = sn − sn−1 , and, given that limn→∞ sn = s, we havelimn→∞ an = limn→∞ (sn − sn−1 ) = limn→∞ sn − limn→∞ sn−1 = s − s = 0.Example 20 The series 1 + q + q 2 + · · · + q n + · · · is often called the geometricseries. Let us investigate its convergence.Since |q n | = |q|n , we have |q n | ≥ 1 when |q| ≥ 1, and in this case the necessarycondition for convergence is not met.Now suppose |q| < 1.
Thensn = 1 + q + · · · + q n−1 =1 − qn1−q1and limn→∞ sn = 1−q, since limn→∞ q n = 0 if |q| < 1.∞ n−1converges if and only if |q| < 1, and in that case itsThus the series n=1 q1sum is 1−q.Example 21 The series 1 + 12 + · · · + n1 + · · · is called the harmonic series, sinceeach term from the second on is the harmonic mean of the two terms on either sideof it (see Exercise 6 at the end of this section).The terms of the series tend to zero, but the sequence of partial sumssn = 1 +11+ ··· + ,2nas was shown in Example 10, diverges. This means that in this case sn → +∞ asn → ∞.Thus the harmonic series diverges.Example 22 The series 1 −1 + 1 − · · · + (−1)n+1 + · · · diverges, as can be seenboth from the sequence of partial sums 1, 0, 1, 0, . . .
and from the fact that the termsdo not tend to zero.If we insert parentheses and consider the new series(1 − 1) + (1 − 1) + · · · ,whose terms are the sums enclosed in parentheses, this new series does converge,and its sum is obviously zero.3.1 The Limit of a Sequence97If we insert the parentheses in a different way and consider the series1 + (−1 + 1) + (−1 + 1) + · · · ,the result is a convergent series with sum 1.If we move all the terms that are equal to −1 in the original series two places tothe right, we obtain the series1 + 1 − 1 + 1 − 1 + 1 − ··· ,we can then, by inserting parentheses, arrive at the series(1 + 1) + (−1 + 1) + (−1 + 1) + · · · ,whose sum equals 2.These observations show that the usual laws for dealing with finite sums can ingeneral not be extended to series.There is nevertheless an important type of series that can be handled exactly likefinite sums, as we shall see below.
These are the so-called absolutely convergentseries. They are the ones we shall mainly work with.b. Absolute Convergence. The Comparison Theorem and Its ConsequencesDefinition 21 The seriesconverges.∞n=1 anis absolutely convergent if the series∞n=1 |an |Since |an + · · · + am | ≤ |an | + · · · |am |, the Cauchy convergence criterion impliesthat an absolutely convergent series converges.The converse of this statement is generally not true, that is, absolute convergenceis a stronger requirement than mere convergence, as one can show by an example.Example 23 The series 1 − 1 + 12 − 12 + 13 − 13 + · · · , whose partial sums are either1n or 0, converges to 0.At the same time, the series of absolute values of its terms1+1+1 1 1 1+ + + + ···2 2 3 3diverges, as follows from the Cauchy convergence criterion, just as in the case ofthe harmonic series: 1111 n + 1 + n + 1 + ··· + n + n + n + n =111=2> 2n ·+ ··· += 1.n+1n+nn+n983LimitsTo learn how to determine whether a series converges absolutely or not, it sufficesto learn how to investigate the convergence of series with nonnegative terms.
Thefollowing theorem holds.Theorem 7 (Criterion for convergence of series of nonnegative terms) A series a1 +· · · + an + · · · whose terms are nonnegative converges if and only if the sequence ofpartial sums is bounded above.Proof This follows from the definition of convergence of a series and the criterionfor convergence of a nondecreasing sequence, which the sequence of partial sumsis, in this case: s1 ≤ s2 ≤ · · · ≤ sn ≤ · · · .This criterion implies the following simple theorem, which is very useful in practice.∞Theorem 8 (Comparison theorem) Let ∞n=1 an andn=1 bn be two series withnonnegative terms. If there exists anindex N ∈ N such that an ≤ bn for alln∞> N ,bimpliestheconvergenceofthen the convergence of the series ∞n=1 nn=1 an ,∞aimpliesthedivergenceofb.and the divergence of ∞nnn=1n=1Proof Since a finite number of terms has no effect on the convergence of a series, we canassume withan ≤ bn for every index n ∈ N.no loss of generality thatThen An = nk=1 ak ≤ nk=1 bk = Bn .
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