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

MATH101 April 2013

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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.

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?

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

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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.

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.

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