MATH 1310.03F A **Exam 3** Thursday, November 30, 2000

NAME:

Student Number: No.Marks

**Instructions:**

1. You have 50 minutes for this exam. You are not allowed to use any calculators.

2. This exam contains 4 questions and has a total of 100 marks.

3. Show all of your work. Your work must justify the answer that you give.

1. (30 Marks)Determine convergence or divergence for each of the
following series.

Solution: (*a*) (Method 1: Bounded Sum Test) Let
. Note
that for . Hence, for ,

Hence, is bounded and converges.

There is another way to show is bounded. Note that for . Hence, for ,

(Method 2: Ordinary Comparison Test) Let , Then for . Since converges, it follows that converges.

Another way: for . Since converges, it follows that converges. (Method 3: Ratio Test) . It follows from Ratio Test that the series converges.

(Method 4: Root Test) Noting that , we have . It follows that converges.

(*b*) Since and
diverges,
it follows from the Limit Comparison Test that diverges.

2. (30 Marks)Determine whether each of the following
integrals series converges absolutely, converges conditionally
or diverges.

Solution. (*a*)
(1) The series
diverges since it is
a *p*-series with *p*=1/2<1.

(2) The series is an alternating series and satifies

(*i*) for all *n*.

(*ii*) .

It follows from the Alternating Serire Test that converges.

(3) By (1) and (2), converges conditionally.

(*b*). Since for each and converges, it follows from the Ordinary
Comparision Test that the series
converges
absolutely.

(20 Marks)
Find the convergence set for

Solution.
Let and
. Then
and

Hence, we have

(*i*) when |*x*-1|<2 (or -1<*x*<3), the series converges absolutely.

(*ii*) When |*x*-1|>2 (or *x*<-1 or *x*>3), the series diverges.

(*iii*) When *x*-1=2 (or *x*=3), the series becomes
and diverges.

(*iv*) When *x*-1=-2 (or *x*=-1), the series becomes
. It follows from
the Alternating Series Test
that converges.

Hence, the convergence set is or [-1, 3).

(20 Marks) Find a power series representation for the
following
function and specify the radius of convergence of the new series

Solution. .
Since

we have

Hence, .

Thu Nov 30 11:19:04 EST 2000