Science:Math Exam Resources/Courses/MATH103/April 2010/Question 01 (d)
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q2 • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q4 (c) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q7 (c) • Q8 •
Question 01 (d) 

Multiple Choice Question: Exactly ONE of the answers provided is correct. There is no partial credit in this question.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

A necessary (but not sufficient) condition that a series converges is that the summands go to zero. Which of the series above don't satisfy this criterion, and hence can not be convergent? 
Hint 2 

For the remaining candidates, use the pseries test. 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. Let's first check if the summands of the offered series go to zero.
Thus iii., iv., and v. can not converge. But the series i. and ii. are candidates for convergent series. By a pairwise comparison of the summands of series i. and series ii. we see that k^{2} < 1/k, and hence (Series i.) < (Series ii.). We are told that only one series converges, so it can only be Series i. (If Series ii. would converge, and (Series i.) < (Series ii.), then Series i. would also converge.) Indeed this is true, by the pseries test with p = 2. Thus, the correct answer is i. Fun fact: 