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# Sequences and series: exercise sheet 3

Exercise 3.1

A sequence (an) is given by

a1=2an+1=an+2an2

(a) Prove that

an+12-2=14an2-22an2

for all n.

(b) Deduce that an>2 for all n.

(c) Using induction, show that

an2-284n

for each n1.

(d) Using an2-2=(an-2)(an+2) deduce that an2 as n.

Exercise 3.2

Suppose in each case that (an) is a sequence whose nth term is given by the expressions below. Prove that (an) converges and find the limit. If you use the continuity of any standard functions in your answer, state which function(s) you use that you know to be continuous and at what value l or (l,m)2 the continuity of that(those) function(s) is required.

(a) n2+2n+12n2+3 ; (b) 1-2n-3n22-3n-4n2 ; (c) (n+1)2-(n-1)2n+1; (d) 3n2n2+n+1+3n2+1.

Exercise 3.3

Prove that the following sequences (an) do not converge, using the following method.

Suppose first that the sequence in question converges to some l. Then find convergent sequences (bn), (cn), ..., and continuous functions f,g, such that some expression involving f,g,,(an),(bn),(cn), evaluates to a sequence that you in fact know to be non-convergent.

The convergence or nonconvergence of any sequences you use must be justified either by stating that they are "standard" convergent/nonconvergent sequences discussed in lectures (such as 1n or (-1)n) or by giving a proof.

(a) (-1)n(1-2-n)+1.

(b) (-1)n(2-n)+n21+n.

(c) 12(cos(n)+(-1)n1n).

(d) sin(n)+ncos(n)1+n.

Exercise 3.4

The sequence (an) is defined inductively by a1=a2=1 and an+2=an+an+1+14. Prove the following by induction on n.

(a) an1 for all n.

(b) an12 for all n.

(c) Prove the following holds for all n1: anan+1an+13.

(d) Say whether (an) has a limit in and if it does, find this limit.

Exercise 3.5

The sequence (cn) is defined by c1=1, c2=12, and cn+2=3cn+1+cn4. Prove the following.

(a) 12cn1 for all n.

(b) cn+1-cn=(-1)n24n for all n.

(c) The subsequences (c2n-1) and (c2n) are both monotonic. (Hint: use (b) above to find a formula for cn+2-cn.)

(d) Explain why these results imply that (c2n-1) and (c2n) both converge, and in fact both converge to the same limit l. Which theorem from the course allows you to deduce that l is the limit of the original sequence (cn)?

(e) Find the limit l. Suggestion: Use (b) repeatedly to get an expression for cn in terms of n. Sum this series then find the limit.

## Sample solutions to selected exercises

Solution to 3.2(a). Write

n2+2n+12n2+3=1+2n+1n22+3n2

and observe that 1n0 and 1n20 as n and so by the constant sequences 1,2 and the continuity of +,, we have 1+2n+1n21 as n. Also by the constant sequences 2,3 and the continuity of +, again, we have 2+3n22 as n. By the continuity of division at (1,2) we have

1+2n+1n22+3n212

Solution to 3.3(a). Suppose an=(-1)n(1-2-n)+1 defines a sequence converging to l. Consider

(an-1)(1-2-n)

Now 1 is a constant sequence, so converges to 1. The sequence 2-n converges to 0 by a result from lectures. So by the continuity of - at (0,1) 1-2-n1-0=1 as n, by the continuity of - at (l,1), (an-1)l-1 as n and by the continuity of division at (l-1,1), the sequence above converges to l-11. But this is impossible as

(an-1)(1-2-n)=(-1)n

which is known from lectures to be nonconvergent.

Solution to 3.4. (a) Let H(n) be the statement an1. Then H(1) is true, as a1=11. Also H(2) is true, as a2=11. Now suppose H(1),H(2),H(n),H(n+1) are all true, where n1 and argue as follows.

an+2=an+an+1+141+1+14=341

using H(n) and H(n+1), which say that an1 and an+11. So H(n+2) is true. This completes the proof by induction.

(Note: If you are not happy about assuming an induction hypothesis that says that H(1),H(2),H(n),H(n+1) are all true, think of this argument just given as showing that P(1) holds and P(n)P(n+1) holds for n where the statement P(n) is k (1kkn+1H(k)). Often induction proofs require the induction hypothesis to cover all previous cases of the induction statement.)

(b) Let H(n) be the statement an12. H(1) is true as a1=112 and H(2) is true as a2=112. Now suppse H(k) is true for all 1kn+1 where n1 and argue as follows.

an+2=an+an+1+141/2+1/2+14=24=12

by H(n) and H(n+1), so H(n+2) is true. This completes the proof by induction.

(c) Let the statement H(n) be anan+1an+13, so as a1=a2=1, H(1) says 1123 which is true. Now assume H(n) is true for n1 and compute:

an+1-an+2=4an+1-an-an+1-14=3an+1-an-14(an+1)-an-140

as an+1an+13. Also,

an+2-an+1+13=3(an+1+an+1)-4(an+1+1)12=3an-an+1-1123an+1-an+1-1120

as anan+1 and an+112. This proves H(n+1). So H(n) is true for all n, by induction.

(d) These results show that (an) is bounded (by (a) and (b)) and is monotonic (by (c)), and so by monotone convergence has a limit l. Now (an+1) and (an+2) are subsequences of (an) so by the theorem on subsequences converge to the same limit l. But an+2=an+an+1+14, so by the continuity of + and division by 4, an+2l+l+14 as n. By uniqueness of limits we have l=l+l+14, and solving this we get l=12. Hence l=12.

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