Supplement: Accumulation Points of Sequences

Example 5.23 The series _∞

k=1k^−0.99 diverges. This follows from the fact that k⁻¹≤k^−0.99. Therefore, the series _∞

k=1k^−0.99 dominates the harmonic series which itself is divergent; see Example5.20.

Example 5.24 In Chap. 2 Euler’s number, e=^∞

j=0

1

j!=1+1+1 2+1

6+ 1 24+ 1

120+ · · ·,

was introduced. We can now show that this definition makes sense, i.e., the series converges. Forj≥4 it is obvious that

j! =1·2·3·4·5·. . .·j≥2·2·2·2·2·. . .·2=2^j. Thus the geometric series_∞

j=0

₁

2

jis a dominating convergent series.

Example 5.25 The decimal notation of a positive real number a=A.α1α2α3. . .

withA∈N0, αk∈ {0, . . . ,9}can be understood as a representation by the series a=A+^∞

k=1

α_k10^−k.

The series converges sinceA+9_∞

k=110^−k is a dominating convergent series.

5.4 Supplement: Accumulation Points of Sequences

Occasionally we need sequences which themselves do not converge but have con-vergent subsequences. The notions of accumulation points, limit superior and limit inferior are connected with this concept.

Definition 5.26 A numberbis called accumulation point of a sequence(a_n)_n≥1if each neighbourhoodU_ε(b)ofbcontains infinitely many terms of the sequence:

∀ε >0∀n∈N∃m=m(n, ε)≥n: |b−a_m|< ε.

Figure5.3displays the sequence an=arctann+cos(nπ/2)+1

nsin(nπ/2).

It has three accumulation points, namely b₁=π/2+1≈2.57,b₂=π/2≈1.57 andb3=π/2−1≈0.57.

58 5 Sequences and Series Fig. 5.3 Accumulation

points of a sequence

If a sequence is convergent with limitathenais the unique accumulation point.

Accumulation points of a sequence can also be characterised with the help of the concept of subsequences.

Definition 5.27 If 1≤n1< n2< n3<· · · is a strictly monotonically increasing sequence of integers (indices), then

(a_nj)_j≥1

is called a subsequence of the sequence(an)n≥1.

Example 5.28 We start with the sequencea_n=¹_n. If we take for instancen_j=j², then we obtain the sequencean_j =_j¹2 as subsequence:

(an)n≥1=

Proposition 5.29 A numberbis an accumulation point of the sequence(a_n)_n≥0if and only ifbis the limit of a convergent subsequence(a_nj)_j≥1.

Proof Letbbe an accumulation point of the sequence(a_n)_n≥0. Step by step we will construct a strictly monotonically increasing sequence of indices(n_j)_j≥1so that

|b−an_j|<1

j (5.2)

is fulfilled for allj∈N. According to Definition5.26forε1=1 we have

∀n∈N∃m≥n: |b−am|< ε1.

We choosen=1 and denote the smallestm≥nwhich fulfils this condition byn₁. Thus

|b−an₁|< ε1=1.

5.4 Supplement: Accumulation Points of Sequences 59 Forε2=¹₂ one again obtains according to Definition5.26:

∀n∈N∃m≥n: |b−a_m|< ε2.

This time we choosen=n1+1 and denote the smallestm≥n1+1 which fulfils this condition byn2. Thus

|b−a_n2|< ε₂=1 2.

It is clear how one has to proceed. Oncen_jis constructed one setsε_j+1=1/(j+1) and uses Definition5.26according to which

∀n∈N∃m≥n: |b−a_m|< ε_j+1.

We choosen=n_j+1 and denote the smallestm≥n_j+1 which fulfils this condi-tion byn_j+1. Thus

|b−an_j+1|< εj+1= 1 j+1.

This procedure guarantees on the one hand that the sequence of indices(n_j)_j≥1is strictly monotonically increasing and on the other hand that the inequality (5.2) is fulfilled for allj∈N. In particular,(a_nj)_j≥1is a subsequence that converges tob.

Conversely, it is obvious that the limit of a convergent subsequence is an

accu-mulation point of the original sequence.

In the proof of the proposition we have used the method of recursive definition of a sequence, namely the subsequence(a_nj)_j≥1.

We next want to show that each bounded sequence has at least one accumulation point—or equivalently—a convergent subsequence. This result bears the names of Bolzano²and Weierstrass³and it is an important technical tool for proofs in many areas of analysis.

Proposition 5.30 (Theorem of Bolzano–Weierstrass) Every bounded sequence (an)n≥1has (at least) one accumulation point.

Proof Due to the boundedness of the sequence there are boundsb < cso that all terms of the sequenceanlie betweenbandc. We bisect the interval[b, c]. Then in at least one of the two half-intervals[b, (b+c)/2]or[(b+c)/2, c]there have to be infinitely many terms of the sequence. We choose such a half-interval and call it[b1, c1]. This interval is also bisected; in one of the two halves again there have to be infinitely many terms of the sequence. We call this quarter-interval[b2, c2].

2B. Bolzano, 1781–1848.

3K. Weierstrass, 1815–1897.

60 5 Sequences and Series Continuing this way we obtain a sequence of intervals[b_n, c_n]of length 2⁻ⁿ(c−b) each of which contains infinitely many terms of the sequence. Obviously thebnare monotonically increasing and bounded, therefore converge to a limitb. Since each interval [b−2⁻ⁿ, b+2⁻ⁿ]by construction contains infinitely many terms of the sequence,bis an accumulation point of the sequence.

If the sequence (a_n)_n≥1 is bounded then the set of its accumulation points is also bounded and hence has a supremum. This supremum is itself an accumulation point of the sequence (which can be shown by constructing a suitable convergent subsequence) and thus forms the largest accumulation point.

Definition 5.31 The largest accumulation point of a bounded sequence is called limit superior and is denoted by limn→∞a_n or lim sup_n→∞a_n. The smallest accu-mulation point is called limit inferior with the corresponding notation lim_n→∞a_nor lim infn→∞an.

follow easily from the definition and justify the notation.

For example, the sequence(a_n)_n≥1from Fig.5.3has lim sup_n→∞a_n=π/2+1 and lim inf_n→∞a_n=π/2−1.

5.5 Exercises

1. Find a law of formation for the sequences below and check for monotonicity, boundedness and convergence:

3. Determine a recursion formula that provides the terms of the geometric se-quence a_n=qⁿ,n≥0 successively. Write aMATLABprogram that calculates the firstN terms of the geometric sequence for an arbitraryq∈R.

5.5 Exercises 61 Check the convergence behaviour for different values ofq and plot the re-sults. Do the same with the help of the applet Sequences.

4. Investigate whether the following sequences converge and, in case of conver-gence, compute the limit:

5. Investigate whether the following sequences have a limit or an accumulation point. Compute, if existent, lim,lim inf,lim sup,inf,sup:

a_n= n+7 6. Open the applet Sequences, visualise the sequences from Exercises 4 and 5 and

discuss their behaviour by means of their graphs.

7. The population model of Verhulst from Example5.3can be described in appro-priate units in simplified form by the recursive relationship

x_n+1=rx_n(1−x_n), n=0,1,2,3, . . .

with an initial valuex0and a parameterr. We presume in this sequence that 0≤x0≤1 and 0≤r≤4 (since allxn then stay in the interval[0,1]). Write aMATLAB-program which calculates for givenr, x0, N the firstN terms of the sequence(xn)n≥1. With the help of your program (and some numerical values forr, x0, N) check the following statements:

(a) For 0≤r≤1 the sequencexnconverges to 0.

(b) For 1< r <2√

2 the sequencexntends to a positive limit.

(c) For 3< r <1+√

6 the sequence xn eventually oscillates between two different positive values.

(d) For 3.75< r≤4 the sequencexnbehaves chaotically.

Illustrate these assertions also with the applet Sequences.

8. The sequence(an)n≥1is given recursively by a1=A, a_n+1=1

2a²_n−1 2.

Which starting valuesA∈Rare fixed points of the recursion, i.e.A=a1= a2= · · ·? Investigate for which starting valuesA∈Rthe sequence converges or diverges, respectively. You can use the applet Sequences. Try to locate the regions of convergence and divergence as precisely as possible.

62 5 Sequences and Series 9. Write aMATLABprogram which, for givenα∈ [0,1]andN∈N, calculates the

firstN terms of the sequence

xn=nα− nα, n=1,2,3, . . . , N

(nαdenotes the largest integer smaller thannα). With the help of your pro-gram, investigate the behaviour of the sequence for a rationalα=^p_q and for an irrationalα(or at least a very precise rational approximation to an irrationalα) by plotting the terms of the sequence and by visualising their distribution in a histogram. Use theMATLABcommandsfloorandhist.

10. Give formal proofs for the remaining rules of calculation of Proposition5.7, i.e., for addition and division by modifying the proof for the multiplication rule.

11. Check the following series for convergence with the help of the comparison criteria: 12. Check the following series for convergence:

∞

13. Try to find out how the partial sumsS_nof the series in Exercises 11 and 12 can be calculated with the help of a recursion and then study their behaviour with the applet Sequences.

14. Prove the convergence of the series ∞

k=0

2^k k!.

Hint. Use the fact thatj! ≥4^j is fulfilled forj ≥9 (why)? From this it follows that 2^j/j! ≤1/2^j. Now apply the appropriate comparison criterion.

15. Prove the ratio test for series with positive termsa_k>0: If there exists a num-berq, 0< q <1 such that the quotients satisfy successivelya_k≤a₀q^kfor allk. Now use the comparison criteria and the con-vergence of the geometric series withq <1.

6