Show by induction Bernouilli’s inequality which states that

[Pages:2]Solutions to the practice problems for midterm I

4. a) Show by induction Bernouilli's inequality which states that

(1 + x)n 1 + nx for x -1 and for all n N.

Solution: Suppose x -1, we call P(n) the statement: (1 + x)n 1 + nx. ?Then P(1) is true since (1 + x)1 = 1 + x. ? Now suppose that P(n) is true for some n N. We must to show that P(n + 1) is true. We know that (1 + x)n+1 = (1 + x)n(1 + x) and because P(n) is true, one gets

(1 + x)n(1 + x) (1 + nx)(1 + x) = 1 + nx + x + nx2.

But, 1 + nx + x + nx2 1 + (n + 1)x since nx2 0. Hence

(1 + x)n+1 1 + (n + 1)x.

? Hence P(n + 1) is true. Consequently, P(n) is true for all n N.

QED

5 b) Find the following limit

lim

n

2n 3n2

+

2n + 3n2

1

+

...

+

3n 3n2

Solution: Notice that the sum 2n + (2n + 1) + (2n + 2) + ? ? ? + 3n contains (n + 1) copies of 2n. So, one has

2n + (2n + 1) + (2n + 2) + ? ? ? + 3n = (n + 1)(2n) + (0 + 1 + 2 + ? ? ? + n)

There follows

=

(n

+

1)(2n)

+

n(n + 2

1)

=

5n(n + 2

1)

lim

n

2n 3n2

+

2n + 3n2

1

+

..

.

+

3n 3n2

= lim

n

5n(n + 1) 6n2

=

5 .

6

6 a) One has:

n-1 5n + 2

n

+ cos(n) 5n + 2

n+1 .

5n + 2

QED

1

Moreover

lim

n

n-1 5n + 2

=

n+1 lim

n 5n + 2

=

1 5

Hence

n lim

n

+ cos( 5n + 2

n)

=

1 5

6 b) (i) for any n N, we have

0

<

b n

=

1

1 + |an|

<

1

Hence the sequence {bn} is bounded. By the Bolzano-Weierstrass Theorem, it has a convergent subsequence.

(ii)(i) for any n N, we have

-1

cos

a n

1

thud

2

cos

a n

4

Thus, the sequence {bn} is bounded. By the Bolzano-Weierstrass Theorem, it has a convergent subsequence.

(iii) We have

b n

=

a n

5

+

a n

=

5

+

a n

-

5

5

+

a n

=

1-

5

5

+

a n

If

we

choose

a n

=

-5 -

1 n

Then

b n

=

1 + n.

This

sequence

is

unbounded

and

it has no convergent subsequence.

2

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download