Then (a) lim n→∞(an+ bn

Calculus 2, Quiz Instructor: Frank Liou

Exam Time: 20 min

Total Score: = min{Your score, 20}.

Remark 1: There is no partial credit if the final answer is wrong.

Remark 2: Indicate the properties of limit given below when you use it in the test. You also need to mention the reason why you can use them.

(Properties of limit) Let (a_n) and (b_n) be sequences such that lim

n→∞a_n= a and lim

n→∞b_n= b.

Then (a) lim

n→∞(an+ bn) = a + b.

(b) lim

n→∞(kan) = ka for k ∈ R.

(c) lim

n→∞(a_nb_n) = ab.

(d) lim

n→∞

a_n b_n = a

b, if b 6= 0.

(e) lim

n→∞a^m_n = a^m where m ≥ 1.

(f) lim

n→∞

m√

an= ^m√

a where m ≥ 1.

(1) (4 Points) Let (a_n), (b_n) and (c_n) be sequence of real numbers. Suppose

n→∞lim an= 1, lim

n→∞bn= −1, lim

n→∞cn= 3.

Evaluate the following limits.

(a) lim

n→∞(2an− 3b_n) = 5 by (a), (b).

(b) lim

n→∞anc³_n= 27 by (c), (e).

(c) lim

n→∞

√cn=

√

3 by (f) (here the limit is well-defined because there exists N > 0 such that cn> 0 for n ≥ N.)

(d) lim

n→∞

anbn

c_n = −1

3 by (c) and (d).

(2) (8 points) Let (a_n) be a sequence of real numbers defined by a_n+1 =√

2 + a_n, n ≥ 1 and a₁=√

2. Show that (a_n) is convergent and evaluate its limit.

Solution: Let us show that (a_n) is bounded above by induction. When n = 1, a1=√

2 < 2. We assume that ak < 2 holds. By induction hypthesis, a_k+1=√

2 + a_k<√

2 + 2 =√ 4 = 2.

1

2

The statement is true for n = k + 1. By induction, the statement is true for all n ≥ 1.

Now let us prove the sequence is increasing. The statement is true for n = 1 : a₂ =

q 2 +√

2 >√

2 = a₁. Suppose ak+1 > ak. Then

ak+2=p2 + a_k+1>√

2 + ak= ak+1.

The statement is true for n = k + 1. By induction, the statement is true for n ≥ 1.

Using monotone sequence properties, (an) is convergent. Let a be its limit. Using property (a) and (f), we see

a = lim

n→∞a_n+1= lim

n→∞

√2 + a_n=√

2 + a > 0.

After solving the equation a =√

2 + a, we obtain a = 2.

(3) (8 Points) Define a sequence of real numbers (a_n) by an= sin 1

2 +sin 2

2² + · · · +sin n

2ⁿ , n ≥ 1.

Show that (a_n) is convergent.

Proof. For n > m,

an− a_m = sin(m + 1)

2^m+1 + · · · + sin n 2ⁿ .

Using triangle inequality and | sin θ| ≤ 1 for all θ ∈ R, we obtain

|a_n− a_m| ≤ 1

2^m+1 + · · · + 1 2ⁿ = 1

2^m



1 − 1 2^n−m



< 1 2^m. Using 2^m > m for all m ≥ 1, we obtain

|a_n− a_m| < 1 m for n > m ≥ 1.

Given  > 0, choose N = [1/] + 1. For every n > m ≥ N,

|a_n− a_m| < 1 m < .

This shows that (an) is a Cauchy sequence. By completeness of R, (an) is conver- gent.

 (4) (Bonus: 10 Points) Suppose (an) is a convergent sequence of real numbers with

n→∞lim a_n= a. Assume a > 0. Show that

n→∞lim

√3

a_n=√³ a.

3

Proof. We know

x³− y³ = (x − y)(x²+ xy + y²).

For each n ≥ 1,

√3

a_n−√³

a = a_n− a

a^2/3n + a^1/3n a^1/3+ a^2/3 . For each n ≥ 1,

|√³

a_n−√³

a| < |a_n− a| 1

√3

a².

Since (an) is convergent to a, for any  > 0, there exists N∈ N so that

|a_n− a| < √³ a.

This implies that for n ≥ N_,

|√³

an−√³

a| < |an− a| 1

√3

a² <

√3

a

√3

a = .

This proves our assertion.