Assignment-1

Solution:

Given any , we must find N sufficiently large so that

for every n>N, .

.

We choose , then n>N, implies that .

Solution :

Exercise 1

Exercise 2

(a) .

So the sequence converges.

(b)So the sequence converges.

(c)

Hence

So the sequence converges.

(d) When n is even number, .

When n is odd number, .So the sequence diverges.

(e) When n is even number, .

When n is odd number, .So the sequence diverges.

(f)

So the sequence converges.

(g)

Hence .

So the sequence converges.

Solution:

(a)

So for all n. The sequence is decreasing.

(b)

So for all n. The sequence is increasing.

Solution :

(a)

So the sequence is bounded by 3.

(b)

So the sequence is bounded by 1/2.

Exercise 3

Exercise 4

a) If the sequence has a finite limit, then the limit is

unique.

Solution:

Suppose the sequence has two different limits a<b.

and .

We choose ., for , there must exist so that for every

, .

, for , there must exist so that for every , .

.

Then , for every implies

.

That’s contradiction.

So the limit is unique. b) If , then is bounded.

Solution:

Set . . For , there exists N>0, such that n>N

Exercise 5

implies that .

Then for all n , we have .

So the sequence is bounded by K.

Solution:

(1) Prove the sequence is increasing.First note that and . It follows that

. So the statement is true for n=1.

Now assume that the statement is true for n=k, that is and show that the statement is true for n=k+1.

Note that .

Thus, by mathematical induction, the sequence is

increasing.

(2) Prove the sequence is bounded by 2.First note that . So the statement is true for n=1.

Now assume that the statement is true for n=k, that is . And show that the statement is true for n=k+1.

Note that . Thus, by mathematical

induction, the sequence is bounded by 2.

Exercise 6

Since is monotonic and bounded, the sequence

converges.Let , then

Since for all n, , we can get the limit of the sequence

is 2.