0.1 Exercises on Sequences

<strong>0.1</strong>. EXERCISES ON SEQUENCES 1
<strong>0.1</strong> <strong>Exercises</strong> on Sequences
In the following exercises find the 1) limit superior and 2) the limit
inferior of the given sequence. Determine whether 3) the limit exists
as n →∞and give reasons.
1. an = (−1)n
n +1
2. an = n (−1) n
does not exist.
Answer: All limits are equal and equal to zero.
Answer: lim sup an =+∞; lim inf
n→∞
n→∞ an = −∞, lim
n→∞ an
3. xn =(−1) n +(−1) n+1 , Answer: All limits are equal and equal to
zero.
4. xn =(−1) n +(−1) n+2 , Answer: lim sup
lim
n→∞
n→∞
xn does not exist.
5. bn = n sin
2
n
xn = +2; lim inf
n→∞ xn = −2,
Answer: All limits are equal and equal to 2.
6. xm =sinmπ +cosmπ Answer: lim sup
lim
m→∞ xm does not exist.
7. an =2(−1) n + n
n +1
m→∞
xm = +1; lim inf
m→∞ xm = −1,
. Answer: lim sup an = +3; lim inf
n→∞
n→∞ an = −1,
lim
n→∞ an does not exist.
8. an =2(−1) n + n
n +1
. Answer: lim sup an = +3; lim inf
n→∞
n→∞ an = −1,
lim
n→∞ an does not exist.
9. xn = sinn Answer: Hard! lim sup xn = +1; lim inf
n→∞
n→∞ xn = −1,
lim
n→∞ xn does not exist. The point here is that π can be approximated
arbitrarily closely by rational numbers (of the form n
m ) so that if we choose
n, m carefully we can make sin n as close to 1 and −1 aswewant1 1 To see this we approximate the number π/2 by rational numbers m/p where p is of the
form 4m +1 (and m is an integer). To do this we have to use a result from Number Theory
(e.g., G.H. Hardy’s, Introduction to the Theory of Numbers) that states: If x is irrational
there exists numbers m, p such that
m
− x
<
p 1
.
p2 So, for each given integer n we choose a pair of numbers mn,pn (with pn →∞as n →∞)
such that
mn
−
pn
π
<
2
1
.
pn
2

2
10. xn =2cosn Answer: Hard! lim sup
lim
n→∞ xn does not exist.
n→∞
11. xn =3xn−1 where n =1, 2, 3,... and x0 = −2.
Answer: All the limits are equal to −∞.
12. xn = −xn−1 where n =1, 2, 3,... and x0 = −1.
xn = +2; lim inf
n→∞ xn = −2,
Answer: lim sup xn = +1; lim inf
n→∞
n→∞ xn = −1, lim
n→∞ xn does not exist.
Prove that the following sequences defined iteratively (or by recursion)
are Cauchy sequences and find their limits. Prove the existence
of the limit using an ε argument.
13. xn = xn−1 where n =1, 2, 3,... and x0 = A.
Answer: The limit is A.
14. xn = 1
2 xn−1, wheren =1, 2, 3,... and x0 =1.
Answer: The limit is zero.
15. xn = 2
5 xn−1 where n =1, 2, 3,... and x0 = −2.
Answer: The limit is zero.
Miscellaneous problems
16. Prove that the sequence (also known as an arithmetic progression) defined
iteratively by xn = a + xn−1, n ≥ 1, where a is a fixed real number, x0 is
given, is a Cauchy sequence if and only if a =0.
Answer: Note that |xn − xm| = |a||n − m| which can be made arbitrarily
small if and only if a =0.
17. Show that the sequence defined iteratively by xn = a · xn−1 where n =
1, 2, 3,... and x0 is a given real number, has the following properties:
• It converges to 0 if |a| < 1
• It converges to x0 if a =1
• It converges to 0 if a ∈ R and x0 =0
• The limit does not exist if a = −1 andx0= 0
• It converges to ∞ if a>1andx0 > 0
From this we get
mn πpn
− <
2
1
.
pn
We claim that the integers mn form a subsequence of the natural numbers such that
sin mn → 1asn →∞. This is because sin mn =sin(mn + pnπ/2 +pnπ/2) = sin(mn −
pnπ/2) cos(pnπ/2) + cos(mn − pnπ/2) sin(pnπ/2) = cos(mn − pnπ/2), since pn is of the form
4m + 1. Now use the Mean Value Theorem: | sin mn − 1| = | cos(mn − pnπ/2) − cos 0| =
| sin ξ||mn − pnπ/2| ≤|mn− pnπ/2| < 1/pn. The result follows upon letting n →∞. A
similar argument is used for the lim inf result.

<strong>0.1</strong>. EXERCISES ON SEQUENCES 3
• It converges to −∞ if a>1andx0 < 0
• The limit does not exist if a ≤−1andx0= 0
Answer: Show that xn = a n x0 for every n ≥ 1 after which all the results
follow.
18. Determine whether or not the sequence, xn defined for each integer n, by
the finite series
xn = arctan 1 + arctan 2 + arctan 3 + ...+arctann,
for n ≥ 1, converges as n →∞!?
Answer: No, the limit cannot exist since arctan n → π/2 asn →∞and
so the Divergence Test for infinite series applies and this result follows.