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

Exercise3.2

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∈ℝ or (l,m)∈ℝ2
the continuity of that(those) function(s) is required.

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

Suppose first that the sequence in question converges to
some l∈ℝ. 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.

Exercise3.4

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) an≤1 for all n.

(b) an≥12 for all n.

(c) Prove the following holds for all n≥1:
an≥an+1≥an+13.

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

Exercise3.5

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

(a) 12≤cn≤1 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∈ℝ. 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 1n→0 and 1n2→0 as
n→∞ and so by the constant sequences 1,2 and the
continuity of +,⋅, we have 1+2n+1n2→1
as n→∞. Also by the constant sequences 2,3 and the
continuity of +,⋅ again, we have
2+3n2→2 as n→∞. By the continuity
of division at (1,2) we have

1+2n+1n22+3n2→12

Solution to 3.3(a). Suppose an=(-1)n(1-2-n)+1
defines a sequence converging to l∈ℝ. 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-n→1-0=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 an≤1.
Then H(1) is true, as a1=1≤1.
Also H(2) is true, as a2=1≤1.
Now suppose H(1),H(2),…H(n),H(n+1) are all true,
where n≥1 and argue as follows.

an+2=an+an+1+14≤1+1+14=34≤1

using H(n) and H(n+1), which say that
an≤1 and an+1≤1.
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∈ℕ
where the statement P(n) is
∀k(1≤k∧k≤n+1⇒H(k)).
Often induction proofs require the induction hypothesis to cover
all previous cases of the induction statement.)

(b) Let H(n) be the statement an≥12. H(1) is true
as a1=1≥12 and H(2) is true
as a2=1≥12. Now suppse H(k) is true for all 1≤k≤n+1
where n≥1 and argue as follows.

an+2=an+an+1+14≥1/2+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
an≥an+1≥an+13,
so as a1=a2=1, H(1) says
1≥1≥23 which is true. Now assume H(n)
is true for n≥1 and compute:

as an≥an+1
and an+1≥12. 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∈ℝ. 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+2→l+l+14
as n→∞. By uniqueness of limits we have
l=l+l+14, and solving this we get l=12.
Hence l=12.