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

NAME:

Student Number:tex2html_wrap_inline257 No.tex2html_wrap_inline259Marks tex2html_wrap_inline261


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.
displaymath271

Solution: (a) (Method 1: Bounded Sum Test) Let tex2html_wrap_inline275. Note that tex2html_wrap_inline277 for tex2html_wrap_inline279. Hence, for tex2html_wrap_inline279,
displaymath283

Hence, tex2html_wrap_inline285 is bounded and tex2html_wrap_inline287 converges.

There is another way to show tex2html_wrap_inline285 is bounded. Note that tex2html_wrap_inline291 for tex2html_wrap_inline279. Hence, for tex2html_wrap_inline279,
displaymath297

(Method 2: Ordinary Comparison Test) Let tex2html_wrap_inline299, Then tex2html_wrap_inline301 for tex2html_wrap_inline279. Since tex2html_wrap_inline305 converges, it follows that tex2html_wrap_inline287 converges.

Another way: tex2html_wrap_inline309 for tex2html_wrap_inline279. Since tex2html_wrap_inline313 converges, it follows that tex2html_wrap_inline287 converges. (Method 3: Ratio Test) tex2html_wrap_inline317. It follows from Ratio Test that the series converges.

(Method 4: Root Test) Noting that tex2html_wrap_inline319 , we have tex2html_wrap_inline321. It follows that tex2html_wrap_inline287 converges.

(b) Since tex2html_wrap_inline327 and tex2html_wrap_inline329 diverges, it follows from the Limit Comparison Test that tex2html_wrap_inline331 diverges.

2. (30 Marks)Determine whether each of the following integrals series converges absolutely, converges conditionally or diverges.
displaymath335

Solution. (a) (1) The series tex2html_wrap_inline341 diverges since it is a p-series with p=1/2<1.

(2) The series tex2html_wrap_inline349 is an alternating series and satifies

(i) tex2html_wrap_inline353 for all n.

(ii) tex2html_wrap_inline359.

It follows from the Alternating Serire Test that tex2html_wrap_inline349 converges.

(3) By (1) and (2), tex2html_wrap_inline349 converges conditionally.

(b). Since tex2html_wrap_inline373 for each tex2html_wrap_inline375 and tex2html_wrap_inline377 converges, it follows from the Ordinary Comparision Test that the series tex2html_wrap_inline379 converges absolutely.

tex2html_wrap_inline381 (20 Marks) Find the convergence set for
displaymath383

Solution. Let tex2html_wrap_inline385 and tex2html_wrap_inline387. Then tex2html_wrap_inline389 and
displaymath391

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 tex2html_wrap_inline413 and diverges.

(iv) When x-1=-2 (or x=-1), the series becomes tex2html_wrap_inline421. It follows from the Alternating Series Test that tex2html_wrap_inline421 converges.

Hence, the convergence set is tex2html_wrap_inline425 or [-1, 3).

tex2html_wrap_inline429 (20 Marks) Find a power series representation for the following function and specify the radius of convergence of the new series
displaymath431

Solution. tex2html_wrap_inline433. Since
displaymath435

we have
displaymath437

Hence, tex2html_wrap_inline439.




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Kunquan Lan
Thu Nov 30 11:19:04 EST 2000