Science:Math Exam Resources/Courses/MATH101/April 2013/Question 12 (c)

MATH101 April 2013

• Q1 (a) • Q1 (b) • Q1 (c) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q8 • Q9 (a) • Q9 (b) • Q10 (a) • Q10 (b) • Q11 • Q12 (a) • Q12 (b) • Q12 (c) •

Other MATH101 Exams

• April 2011 • April 2010 • April 2009 • April 2008 • April 2007 • April 2005 • April 2006 • April 2012 • April 2013 • April 2014 • April 2015 • April 2016 • April 2018 • April 2017 •

[hide] Question 12 (c)
Full-Solution Problems. In questions 4–12, justify your answers and show all your work. If a box is provided, write your final answer there. Unless otherwise indicated, simplification of numerical answers is required in these questions. Determine, with explanation, whether $\displaystyle \sum _{n=2}^{\infty }{\frac {n+\sin n}{1+n^{2}}}$ converges or diverges.

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.

[show] Hint 1
By part (b), we cannot use integral test. It's clear geometric series, ratio test, and alternating series test do not apply. What's left?

[show] Hint 2
Try either limit comparison test, or the direct comparison 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.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

[show] Solution 1
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Limit comparison test finishes this problem fairly quickly. By inspection, the summand looks like $\displaystyle 1/n$ for large n. Let's prove this rigorously. $\displaystyle \lim _{n\to \infty }{\frac {n+\sin n}{1+n^{2}}}\cdot {\frac {n}{1}}=\lim _{n\to \infty }{\frac {1+{\tfrac {\sin n}{n}}}{{\tfrac {1}{n^{2}}}+1}}=1>0$ and finite. Here, we used $\displaystyle -1/n\leq (\sin n)/n\leq 1/n$ and both the left and right sides limit to 0, so squeeze theorem gives $\displaystyle \lim _{n\to \infty }(\sin n)/n=0$ . Since $\displaystyle \sum _{n=2}^{\infty }{\frac {1}{n}}$ diverges by p-series, we conclude $\displaystyle \sum _{n=2}^{\infty }{\frac {n+\sin n}{1+n^{2}}}$ diverges as well by the limit comparison test.

[show] Solution 2
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We can prove this using the comparison test, in the same way as part (a), with $\displaystyle n$ replacing $\displaystyle x$ and sums instead of integrals. Notice that for large $\displaystyle x$ , the summands behaves like $\displaystyle 1/n$ , so this will be our goal. First, since $\displaystyle \sin n\geq -1$ , we have $\displaystyle {\frac {n+\sin n}{1+n^{2}}}\geq {\frac {n-1}{1+n^{2}}}$ Now, for $\displaystyle n\geq 2$ , we have $\displaystyle {\frac {n-1}{1+n^{2}}}\geq {\frac {1}{3n}}$ To prove this, notice if $\displaystyle n\geq 2$ , then $\displaystyle {\begin{aligned}n(2n-3)&\geq 2\cdot 1\geq 1\\3n^{2}-3x&\geq n^{2}+1\\{\frac {n-1}{1+n^{2}}}&\geq {\frac {1}{3n}}\end{aligned}}$ where in the second line we added $\displaystyle n^{2}$ to both sides, and in the third line we divided by $\displaystyle (3n)(1+n^{2})>0$ . Now, since $\displaystyle \sum _{n=2}^{\infty }{\frac {1}{3n}}={\frac {1}{3}}\sum _{n=2}^{\infty }{\frac {1}{n}}$ diverges by p-series, we have that $\displaystyle \sum _{n=2}^{\infty }{\frac {n+\sin n}{1+n^{2}}}$ diverges by the comparison test.