Science:Math Exam Resources/Courses/MATH105/April 2015/Question 05 (c)

MATH105 April 2015

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Question 05 (c)
Determine whether the series $\sum \limits _{m=1}^{\infty }{\frac {3m+\sin {\sqrt {m}}}{m^{2}}}$ 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
Apply the comparison and p-series tests.

Hint 2
Use the fact that $\sin {\sqrt {m}}\geq -1$ for all m to help you find a series to compare the given one to.

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.

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. We will apply the comparison test to show that the series diverges. The comparison test states that if $a_{n}$ and $b_{n}$ are sequences with $0\leq a_{n}\leq b_{n}$ for $n\geq k$, then if $\sum _{n=k}^{\infty }a_{n}$ diverges, the series $\sum _{n=k}^{\infty }b_{n}$ diverges as well. Since $\sin {\sqrt {m}}\geq -1$ for all $m\geq 1$, $3m+\sin {\sqrt {m}}\geq 3m-1$. Furthermore, $3m-1\geq 2m$ for all $m\geq 1$, so ${\frac {3m+\sin {\sqrt {m}}}{m^{2}}}\geq {\frac {3m-1}{m^{2}}}\geq {\frac {2m}{m^{2}}}={\frac {2}{m}}.$ As $\sum _{m=1}^{\infty }{\frac {2}{m}}$ diverges by the p-series test (here, $p=1$), then by the comparison test, the series $\sum _{m=1}^{\infty }{\frac {3m+\sin {\sqrt {m}}}{m^{2}}}$ diverges as well.

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 use the limit comparison test to show that the series diverges. This test says that if $\{a_{n}\},\{b_{n}\}$ are sequences with $a_{n},b_{n}\geq 0$, and if $0<\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}<\infty$ (and the limit exists), then the series $\sum _{n=k}^{\infty }a_{n}$ and $\sum _{n=k}^{\infty }b_{n}$ either both converge or both diverge. Applying the test with $a_{n}={\frac {3n+\sin {\sqrt {n}}}{n^{2}}}$ and $b_{n}={\frac {1}{n}}$, we observe that $\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\lim _{n\to \infty }{\frac {3n+\sin {\sqrt {n}}}{n^{2}}}\cdot n=\lim _{n\to \infty }{\frac {3n+\sin {\sqrt {n}}}{n}}=\lim _{n\to \infty }3+{\frac {\sin {\sqrt {n}}}{n}}=3,$ which clearly lies between $0$ and $\infty$. As $\sum _{m=2}^{\infty }{\frac {1}{m}}$ diverges (by the p-series test with $p=1$), the series $\sum _{m=2}^{\infty }{\frac {3m+\sin {\sqrt {m}}}{m^{2}}}$ also diverges.