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

MATH105 April 2015

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Question 05 (b)
Determine whether the series $\sum \limits _{n=2}^{\infty }{\frac {n^{2}+n+1}{n^{5}-n}}$ 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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Apply the comparison and p-series tests.

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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. We will apply the comparison test to show that the series converges. The comparison test states that if $a_{n}$ and $b_{n}$ are two sequences with $0\leq b_{n}\leq a_{n}$ for $n\geq k$, then if $\sum _{n=k}^{\infty }a_{n}$ converges, the series $\sum _{n=k}^{\infty }b_{n}$ converges as well. Since $n\geq 2$, $1\leq n^{2}$ and $n\leq n^{2}$. Hence $n^{2}+n+1\leq n^{2}+n^{2}+n^{2}=3n^{2}$. Furthermore, since $n\leq {\frac {1}{2}}n^{5}$ for all $n\geq 2$, we have $n^{5}-n\geq {\frac {1}{2}}n^{5}$. Therefore, ${\frac {n^{2}+n+1}{n^{5}-n}}\leq {\frac {3n^{2}}{{\frac {1}{2}}n^{5}}}={\frac {6}{n^{3}}}.$ Since $\sum _{n=2}^{\infty }{\frac {6}{n^{3}}}$ converges by the p-series test (here, $p=3$), by comparison the series $\sum _{n=2}^{\infty }{\frac {n^{2}+n+1}{n^{5}-n}}$ also converges.

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 converges. 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 $a_{n}={\frac {n^{2}+n+1}{n^{5}-n}}$ and $b_{n}={\frac {1}{n^{3}}}$, we observe that $\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\lim _{n\to \infty }{\frac {n^{2}+n+1}{n^{5}-n}}\cdot n^{3}=\lim _{n\to \infty }{\frac {n^{5}+n^{4}+n^{3}}{n^{5}-n}}=1,$ which clearly lies between $0$ and $\infty$. As $\sum _{n=2}^{\infty }{\frac {1}{n^{3}}}$ converges (by the p-series test with $p=3$), the series $\sum _{n=2}^{\infty }{\frac {n^{2}+n+1}{n^{5}-n}}$ also converges.