Show that the sequence is monotone and bounded

(1) Let x1∈ R with x1> 1. Define (xn) by xn+1= 2 − 1

x_n for n ∈ N.

Show that the sequence is monotone and bounded. What is its limit?

Proof. For n = 1,

x2− x1= 2 − 1 x₁ − x1. Since x1> 1, by A.G. inequality,

x1+ 1 x1

> 2 r

x1· 1 x1

= 2.

Hence x2− x1< 0. Assume that xk+1< xk. Then 1/xk+1> 1/xk and hence xk+2= 2 − 1

x_k+1 < 2 − 1

x_k = xk+1.

We see that xn+1< xn is true for n = k + 1. By mathematical induction, xn+1< xn is true for any n ≥ 1.

Since x1> 1,

x2= 2 − 1 x1

> 2 −1 1 = 1.

Let us claim that x_n> 1 for any n ≥ 1. Suppose the statement is true for n = k. Then xk+1= 2 − 1

xk

> 2 −1 1 = 1.

The statement is true for n = k + 1. By mathematical induction x_n> 1 holds for n ∈ N.

The sequence is convergent in R. Denote its limit by x. Then x = lim

n→∞xn+1= lim

n→∞

 2 − 1

xn



= 2 − 1 x.

We find x = 1. 

(2) Let (a_n) be a sequence in R^p and a₀∈ R^p. Assume that there exists 0 < ρ < 1 such that kan+1− ank ≤ ρkan− an−1k for any n ≥ 1.

Prove that (an) is convergent in R^p.

Proof. For n = 1, we have ka₂− a₁k ≤ ρka₁− a₀k. For n = 2, we have ka3− a2k ≤ ρka2− a1k ≤ ρ²ka1− a0k = ρ³ka1− a0k

ρ .

Take K = ka₁− a₀k/ρ. By induction, we can show that kan− an−1k ≤ ρⁿK.

For n > m ≥ N, using triangle inequality,

kan− amk = k(an− an−1) + · · · + (am+1− am)k

≤ (ρⁿ+ · · · + ρ^m+1)K

= 1 − ρ^n−m 1 − ρ ρ^m+1K

≤ ρ^m+1K 1 − ρ .

1

Since 0 < ρ < 1, lim

m→∞(ρ^mK/(1 − ρ)) = 0. For  > 0, we can choose N ∈ N so that ρ^m+1K/(1 − ρ) <  whenever m ≥ N. (You should be able to find such an Nby yourself.) This implies that kan− amk <  whenever n, m ≥ N. This shows that (an) is a Cauchy sequence in R^p. By completeness of R^p, (an) is convergent.

 (3) Let a ≥ 1 and 0 < z1< 1 +√

1 + 4a

2 . Define a sequence (zn) by zn+1=√

a + zn, for n ∈ N.

Show that (z_n) is convergent and find its limit. Use two different methods to prove the convergence of (z_n).

(a) Use mathematical induction to show that (z_n) is increasing and bounded above. (A sequence (a_n) is bounded above if there exists U ∈ R so that an≤ U for all n ∈ N.) Proof. Using the assumption, we can show that z2> z1. That the sequence is increasing can be shown by induction. Assuming that zk+1> zk, we have

zk+2=pa + zk+1>√

a + zk = zk+1. By assumption, z1< ¹⁺

√1+4a

2 . We assume that zk< ¹⁺

√1+4a

2 is true. Then z_k+1² = a + zk < 1 +√

1 + 4a + 2a

2 = 1 + 2√

1 + 4a + 1 + 4a

4 = 1 +√

1 + 4a 2

² . This implies that zk+1< ¹⁺

√1+4a 2 .

 (b) Use the following observation:

zn+1− zn=√

a + zn−pa + zn−1= zn− zn−1

√a + zn+√

a + zn−1

and for any n ∈ N,

0 < 1

√a + zn+√

a + zn−1

≤1 2. Use exercise (2).

Proof. We obtain that

|zn+1− zn| ≤ 1

2|zn− zn−1| for all n ≥ 2. Then use (2).

 (4) Determine the convergence/divergence of the sequence (xn) defined by

xn= 1

n + 1+ 1

n + 2+ · · · + 1 2n =

n

X

i=1

1 n + i. (5) Let (x_n) be a sequence in R^p.

(a) If (x_n) is convergent to x ∈ R^p, show that lim_n→∞kx_nk = kxk.

Proof. This can be shown by the triangle inequality:

|kxnk − kxk| ≤ kxn− xk.

 (b) If limn→∞kxnk = 0, show that limn→∞xn= 0.

Proof. For any  > 0, there exists N ∈ N such that kxnk = |kxnk − 0| <  whenever n ≥ N. This proves the result.

 (c) If lim_n→∞kx_nk exists, is the sequence (x_n) convergent? If yes, prove it. If not, give a

counterexample.

Proof. Not true. Take xn = (−1)ⁿ. Then (xn) is divergent. Notice that |xn| = 1 for all n ≥ 1 and any constant sequence is convergent. Therefore (|xn|) is convergent while

(x_n) is divergent. 

(6) Assume that lim

n→∞(kxn+1k/kxnk) = r with r < 1.

(a) Show that there exist positive real numbers ρ and C and a natural number N with r < ρ < 1 so that

0 < kxnk < Cρⁿ for n ≥ N.

Proof. Let xn= kxnk. Choose  > 0 so that ρ = r+ < 1. Then there exists N = N∈ N so that |(xn+1/xn) − r| <  for n ≥ N. This implies that 0 < xn+1< ρxn for all n ≥ N.

Thus 0 < x_{N +1}< ρx_N and 0 < x_{N +2}< ρx_{N +1}< ρ²x_N. Inductively, we obtain 0 < x_{N +k}< ρ^kx_N = ρ^{N +k}x_N

ρ^N.

Let C = xN/ρ^N. We find 0 < xn< Cρⁿ for any n ≥ N + 1. This proves the result.

 (b) Show that lim

n→∞x_n= 0.

Proof. By exercise 6a, limn→∞kxnk = 0. By exercise 5b, limn→∞xn= 0.

 (7) Let (an) be a sequence in R^p such that

n→∞lim

pkan nk = r with r > 1.

Prove that (a_n) is unbounded. (Hint: estimate ka_nk_R^p and use the Bernoulli’s inequality:

for x > 0, we have (1 + x)ⁿ ≥ 1 + nx for any n ∈ N.)

Proof. Choose  > 0 so that r −  > 1. Denote ρ = r − . There exists N ∈ N so that

|pkaⁿ _nk − r| <  for n ≥ N. Thus ka_nk > ρⁿ for any n ≥ N. Write ρ = 1 + x for x > 0. By Bernoulli’s inequality,

kank > 1 + nx.

Use this inequality to show that (a_n) is unbounded.

 (8) Let (an) be a sequence in R^p and C > 0.

(a) Assume that

ka_n+1− a_nk ≤ C

n(n + 1), n ≥ 1.

Prove that (a_n) is convergent in R^p. (Hint: show that (a_n) is a Cauchy sequence in R^p and use the completeness of R^p.)

Proof. For n > m, kan− amk ≤ C

 1

m(m + 1)+ · · · + 1 (n − 1)n



< C 1 m. Then the rest is left to the readers.



(b) Let (xn) be a sequence of real numbers defined by x_n+1= x_n+ (−1)ⁿ

(2n + 1)!

with x1= 1. Test the convergence/divergence of (xn). (Hint: use (a)).

Proof. For n ≥ 1,

|xn+1− xn| = 1

(2n + 1)!< 1 2n(2n + 1). Use similar method proved in (a).

 (9) Let Q be the subset of all rational numbers and

d(x, y) = |x − y|, x, y ∈ Q.

Then (Q, d) is a metric space. In this exercise, you will learn that (Q, d) is not a complete metric space.

(a) For each n, define

x_n = 1 + 1

1!+ · · · + 1

n!, n ≥ 1.

Prove that x_n∈ Q for all n ≥ 1.

Proof. Finite sum of rational numbers is rational.

 (b) Prove that (xn) is a Cauchy sequence in (Q, d).

Proof. For n ≥ 1,

|xn+1− xn| = 1

(n + 1)! < 1 n(n + 1). By Exercise 8, (x_n) is a Cauchy sequence in (Q, d).

 (c) Prove that (xn) is divergent in (Q, d). Hint: Suppose (xⁿ) is convergent to some x = p/q in Q. Define pⁿ= n!xnfor n ≥ 1. Show that |n!x−pn| > 0 for n ≥ 1. Estimate |n!x−pn|.

Then use the suggested exercises below to show that such x does not exist.

Proof. Suppose that (xn) is convergent to x = p/q for some p, q ∈ Z with q 6= 0. Assume that q > 0. For m > n,

|xm− xn| = 1

(n + 1)!+ · · · + 1 m!. Multiplying the above equation by n!, we find

|n!x_m− p_n| = 1

n + 1+ · · · + 1

(n + 1)(n + 2) · · · m. For any k ≥ 1,

(n + 1)(n + 2) · · · (n + k + 1) ≥ (n + k)(n + k + 1) In other words,

1

(n + 1)(n + 2) · · · (n + k + 1) ≤ 1

(n + k)(n + k + 1) = 1

n + k − 1 n + k + 1 for any k ≥ 1. This shows that

1

(n + 1)(n + 2)+ · · · + 1

(n + 1)(n + 2) · · · m ≤ 1 n + 1− 1

m.

Therefore for m > n,

|n!x_m− p_n| < 2 n + 1− 1

m < 2 n + 1.

This shows that |n!xm− pn| < 2/(n + 1) for any m > n ≥ 1. By taking m → ∞, we find

|n!x − pn| ≤ 2 n + 1. On the other hand,

|n!xm− pn| = 1

n + 1+ · · · + 1

(n + 1)(n + 2) · · · m > 1 n + 1. By taking m → ∞, we find

|n!x − pn| ≥ 1 n + 1> 0.

We find that

0 < |n!x − p_n| ≤ 2 n + 1 for any n ≥ 1. Multiplying the inequality by q, then

0 < |n!p − qp_n| ≤ 2q n + 1.

By Sandwich principle, limn→∞|n!p − qpn| = 0. Notice that (n!p − qpn) is a sequence of integers. If it is convergent to 0, there exists N ∈ N so that n!p − qpⁿ= 0 for all n ≥ N.

In this case, we find n!x − pn = 0 which leads to the contradiction to |n!x − pn| > 0.

 To do (9), you need to recall the exercise from hw 1 and hw 2 in Calculus I: (you do not need to turn in the following exercises) If you are not familiar with the exercises below, I suggest that you should solve these.

(1) Let (a_n) be a sequence of integers. Suppose that (a_n) is convergent in R. Show that there exists N such that a_n= a_N for any n ≥ N.

(2) Let α be a real number. Suppose that there exist sequence of integers (p_n) and (q_n) and a sequence of positive real numbers (rn) such that

(a) limn→∞rn = 0.

(b) 0 < |pn− αqn| < rn for any n ≥ 1.

Show that α must be an irrational number.