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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