# Mathematical Analysis

Mathematical Analysis

Examples Sheet 8

1. Evaluate X8

n=1

5 – 3

n

2

3n

.

2. Prove that, for n > 2,

1 +

1

n

n

2

>

n

2 – 1

2

. Deduce that X8

n=1

1 +

1

n

-n

2

converges.

3. Show that

log( 3

2

) 6

X8

n=1

1

n(2n + 1) 6

1

3 + log( 3

2

).

4. Let (bn)n?N+ be a bounded sequence. Prove that if P8

n=1 an is absolutely convergent then P8

n=1 anbn

is also absolutely convergent.

5. Sketch the graph of f(x) = v

x, x > 0, and use it, together with relevant calculations, to explain why

2

3N

3/2 <

X

N

n=1

v

n < 2

3

(N

3/2 – 1) + v

N.

6. Find the value of X8

n=2

1

n3 – n

by the telescoping series method.

7. Recall from Question 1 on Examples Sheet 0 that, ? a, b ? R,

|ab| 6

1

2

(a

2 + b

2

). (1)

(a) Suppose that an > 0 ? n ? N+ and that P8

n=1 an < +8. Make appropriate choices of a and b in

(1) to prove that

X8

n=1

v

an

n

< +8.

(b) Provide an example of a sequence (an)n?N+

for which an > 0 ? n ? N+ and P8

n=1

P

an < +8 but

8

n=1 v

an = +8.

Worked Example

Find the value of X8

n=1

v

n + 2 –

v

n

p

n(n + 2)

by the telescoping series method.

Solution

v

n + 2 –

v

n

p

n(n + 2)

=

1

v

n

–

1

v

n + 2

. Therefore,

v

3 –

v

1

v

3

=

1

1

–

1

v

3

v

4 –

v

2

v

8

=

1

v

2

–

1

v

4

v

5 –

v

3

v

15

=

1

v

3

–

1

v

5

.

.

.

.

.

.

v

k + 1 –

v

k – 1

p

(k – 1)(k + 1)

=

1

v

k – 1

–

1

v

k + 1

v

k + 2 –

v

k

p

k(k + 2)

=

1

v

k

–

1

v

k + 2

.

Therefore,

X

k

n=1

v

n + 2 –

v

n

p

n(n + 2)

= 1 +

1

v

2

–

1

v

k + 1

–

1

v

k + 2

.

It follows that

X8

n=1

v

n + 2 –

v

n

p

n(n + 2)

= lim

k?8

1 +

1

v

2

–

1

v

k + 1

–

1

v

k + 2

= 1 +

1

v

2

.

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