Recursively defined sequences

In document Basic Analysis: Introduction to Real Analysis (Page 61-67)

Sequences and Series

2.2.3 Recursively defined sequences

Now that we know we can interchange limits and algebraic operations, we can compute the limits of many sequences. One such class are recursively defined sequences, that is, sequences where the next number in the sequence is computed using a formula from a fixed number of preceding elements in the sequence.

Example 2.2.8: Let{xn}be defined byx1:=2 and xn+1:=xn−x

2 n−2 2xn .

We must first find out if this sequence is well-defined; we must show we never divide by zero. Then we must find out if the sequence converges. Only then can we attempt to find the limit.

So let us prove xn exists and xn>0 for all n(so the sequence is well-defined and bounded

below). Let us show this by . We know thatx1=2>0. For the induction step, suppose xn>0. Then xn+1=xn−x 2 n−2 2xn = 2x2nx2n+2 2xn = x2n+2 2xn . It is always true thatx2

n+2>0, and asxn>0, then x

2 n+2

2xn >0 and hencexn+1>0.

Next let us show that the sequence is monotone decreasing. If we show thatx2n−2≥0 for alln, thenxn+1≤xnfor alln. Obviouslyx12−2=4−2=2>0. For an arbitrarynwe have

x2n+1−2= x2 n+2 2xn 2 −2=x 4 n+4x2n+4−8x2n 4x2 n = x4 n−4x2n+4 4x2 n = x2 n−22 4x2 n .

Since any square is nonnegative,xn2+1−2≥0 for alln. Therefore,{xn}is monotone decreasing and bounded (xn>0 for alln), and so the limit exists. It remains to find the limit.

Write

2xnxn+1=x2n+2.

Since{xn+1}is the 1-tail of{xn}, it converges to the same limit. Let us definex:=limxn. Take the limit of both sides to obtain

2x2=x2+2,

orx2=2. Asxn>0 for allnwe getx≥0, and thereforex=√2.

You may have seen the above sequence before. It is theNewton’s method for finding the square root of 2. This method comes up very often in practice and converges very rapidly. We used the fact thatx21−2>0, although it was not strictly needed to show convergence by considering a tail of the

sequence. In fact the sequence converges as long asx16=0, although with a negativex1we would arrive atx=√2. By replacing the 2 in the numerator we obtain the square root of any positive number. These statements are left as an exercise.

You should, however, be careful. Before taking any limits, you must make sure the sequence converges. Let us see an example.

Example 2.2.9: Supposex1:=1 andxn+1:=x2n+xn. If we blindly assumed that the limit exists (call itx), then we would get the equationx=x2+x, from which we might concludex=0. However, it is not hard to show that{xn}is unbounded and therefore does not converge.

The thing to notice in this example is that the method still works, but it depends on the initial valuex1. If we setx1:=0, then the sequence converges and the limit really is 0. An entire branch of mathematics, called dynamics, deals precisely with these issues. See .

Some convergence tests

It is not always necessary to go back to the definition of convergence to prove that a sequence is convergent. We first give a simple convergence test. The main idea is that{xn}converges toxif and only if{|xn−x|}converges to zero.

Proposition 2.2.10. Let{xn}be a sequence. Suppose there is an xRand a convergent sequence {an}such that

lim

n→∞an=0 and

|xn−x| ≤an for all n. Then{xn}converges andlimxn=x.

Proof. Letε >0 be given. Note thatan≥0 for alln. Find anM∈Nsuch that for alln≥M we

havean=|an−0|<ε. Then, for alln≥Mwe have

|xn−x| ≤an<ε.

As the proposition shows, to study when a sequence has a limit is the same as studying when another sequence goes to zero. In general, it may be hard to decide if a sequence converges, but for certain sequences there exist easy to apply tests that tell us if the sequence converges or not. Let us see one such test. First let us compute the limit of a very specific sequence.

Proposition 2.2.11. Let c>0.

(i) If c<1, then

lim n→∞c

n=0. (ii) If c>1, then{cn}is unbounded.

Proof. First supposec<1. Asc>0, thencn>0 for allnNby . Asc<1, thencn+1<cn

for alln. So we have a decreasing sequence that is bounded below. Hence, it is convergent. Let L:=limcn. The 1-tail{cn+1}also converges toL. Taking the limit of both sides ofcn+1=c·cn, we obtainL=cL, or(1c)L=0. It follows thatL=0 asc6=1.

Now supposec>1. Suppose for contradiction that the sequence is bounded above byB>0,

that is,cnBfor alln

N. Then for alln,

1 c n = 1 cn ≥ 1 B >0.

If we look at the above proposition, we note that the ratio of the(n+1)th term and thenth term isc. We generalize this simple result to a larger class of sequences. The following lemma will come up again once we get to series.

Lemma 2.2.12(Ratio test for sequences). Let{xn}be a sequence such that xn6=0for all n and such that the limit

L:= lim n→∞

|xn+1|

|xn| exists. (i) If L<1, then{xn}converges andlimxn=0.

(ii) If L>1, then{xn}is unbounded (hence diverges).

IfLexists, butL=1, the lemma says nothing. We cannot make any conclusion based on that information alone. For example, the sequence {1/n}converges to zero, butL=1. The constant sequence{1}converges to 1, not zero, andL=1. The sequence{(1)n}does not converge at all, andL=1 as well. Finally, the sequence{n}is unbounded, yet againL=1. The statement may be strengthened, see exercises and .

Proof. SupposeL<1. As |xn+1|

|xn| ≥0 for alln, thenL≥0. Pickrsuch thatL<r<1. We wish to compare the sequence to the sequencern. The idea is that while the sequence is not going to be less thanLeventually, it will eventually be less thanr, which is still less than 1. The intuitive idea of the proof is illustrated in .

1

L r

Figure 2.4:Proof of ratio test in picture. The short lines represent the ratios |xn+1|

|xn| approachingL.

Asr−L>0, there exists anM∈Nsuch that for alln≥Mwe have |xn+1| |xn| −L <r−L Therefore, fornM, |xn+1| |xn| −L<r−L or |xn+1| |xn| <r. Forn>M(that is fornM+1) write

|xn|=|xM||xM+1| |xM| |xM+2| |xM+1|··· |xn| |xn−1| <|xM|rr···r=|xM|r n−M = (|x M|r−M)rn.

The sequence{rn}converges to zero and hence|xM|r−Mrnconverges to zero. By , theM-tail of{xn}converges to zero and therefore{xn}converges to zero.

Now supposeL>1. Pickrsuch that 1<r<L. AsLr>0, there exists anMNsuch that

for alln≥M |xn+1| |xn| −L <L−r.

Therefore,

|xn+1| |xn| >r. Again forn>M, write

|xn|=|xM||xM+1| |xM| |xM+2| |xM+1|··· |xn| |xn−1| >|xM|rr···r=|xM|r n−M = (|x M|r−M)rn.

The sequence{rn}is unbounded (sincer>1), and therefore{xn}cannot be bounded (if|xn| ≤B

for alln, thenrn< B |xM|r

M for alln, which is impossible). Consequently,{xn}cannot converge. Example 2.2.13: A simple application of the above lemma is to prove

lim n→∞ 2n n! =0. Proof: Compute 2n+1/(n+1)! 2n/n! = 2n+1 2n n! (n+1)!= 2 n+1.

It is not hard to see that{n+12 }converges to zero. The conclusion follows by the lemma. Example 2.2.14: A more complicated (and useful) application of the ratio test is to prove

lim n→∞n

1/n=1. Proof: Letε>0 be given. Consider the sequence(1+n

ε)n . Compute (n+1)/(1+ε)n+1 n/(1+ε)n = n+1 n 1 1+ε.

The limit of n+1n =1+1n asn∞is 1, and so lim n→∞ (n+1)/(1+ε)n+1 n/(1+ε)n = 1 1+ε <1. Therefore, n

(1+ε)n converges to 0. In particular, there exists anNsuch that forn≥N, we have

n

(1+ε)n <1, or n<(1+ε)

n, orn1/n<1+ε. Asn1, thenn1/n1, and so 0n1/n1<ε. Consequently, limn1/n=1.

Exercises

Exercise2.2.1: Prove . Hint: Use constant sequences and . Exercise2.2.2: Prove part of .

Exercise2.2.3: Prove that if{xn}is a convergent sequence, k∈N, then lim n→∞x k n= lim n→∞xn k . Hint: Use .

Exercise2.2.4: Suppose x1:=12 and xn+1:=x2n. Show that{xn}converges and findlimxn. Hint: You cannot

divide by zero!

Exercise2.2.5: Let xn:=n−cosn(n). Use the to show that{xn}converges and find the limit.

Exercise2.2.6: Let xn:=n12 and yn:=1n. Define zn:=yxnn and wn:=yxnn. Do{zn}and{wn}converge? What

are the limits? Can you apply ? Why or why not?

Exercise2.2.7: True or false, prove or find a counterexample. If{xn}is a sequence such that{xn2}converges,

then{xn}converges.

Exercise2.2.8: Show that

lim

n→∞

n2 2n =0.

Exercise2.2.9: Suppose{xn}is a sequence and suppose for some x∈R, the limit

L:= lim

n→∞

|xn+1−x|

|xn−x|

exists and L<1. Show that{xn}converges to x.

Exercise2.2.10(Challenging): Let{xn}be a convergent sequence such that xn≥0and k∈N. Then

lim n→∞x 1/k n = lim n→∞xn 1/k .

Hint: Find an expression q such that x1n/k−x1/k

xn−x =

1

q.

Exercise2.2.11: Let r>0. Show that starting with any x16=0, the sequence defined by xn+1:=xn−x 2 n−r 2xn converges to√r if x1>0and−√r if x1<0. Exercise2.2.12:

a) Suppose{an}is a bounded sequence and{bn}is a sequence converging to 0. Show that{anbn}converges

to 0.

b) Find an example where{an}is unbounded,{bn}converges to 0, and{anbn}is not convergent.

Exercise2.2.13(Easy): Prove the following stronger version of , the ratio test. Suppose{xn}

is a sequence such that xn6=0for all n.

a) Prove that if there exists an r<1and MNsuch that for all n≥M we have

|xn+1|

|xn| ≤r,

then{xn}converges to0.

b) Prove that if there exists an r>1and MNsuch that for all n≥M we have

|xn+1|

|xn| ≥r,

then{xn}is unbounded.

Exercise2.2.14: Suppose x1:=c and xn+1:=xn2+xn. Show that{xn}converges if and only if−1≤c≤0,

in which case it converges to 0.

Exercise2.2.15: Prove lim

n→∞(n

2+1)1/n

=1.

Exercise2.2.16: Prove that

(n!)1/n is unbounded. Hint: Show thatCn

Limit superior, limit inferior, and Bolzano–Weierstrass

In document Basic Analysis: Introduction to Real Analysis (Page 61-67)