Comparison Test

Section: 
9.4
Date: 
Tuesday, October 5, 2010 - 14:00 - Wednesday, October 6, 2010 - 15:00
AttachmentSize
fall2010math2300_comparison-test.pdf46 KB

Math 2300 Section 005 – Calculus II – Fall 2010

9.4 Comparison Test Examples – October 5, 2010

  1. Use the comparison test to determine if the given series converges or diverges.

    \sum _{{n=1}}^{\infty}\frac{1}{n^{4}+1}

    Solution: Since

    0\le\frac{1}{n^{4}+1}\le\frac{1}{n^{4}}

    and we know the p-series

    \sum _{{n=1}}^{\infty}\frac{1}{n^{4}}

    converges, the given series also converges.

  2. Use the comparison test to determine if the given series converges or diverges.

    \sum _{{n=1}}^{\infty}\frac{n-2}{n^{3}+2}

    Solution: We examine instead the convergence of

    \sum _{{n=2}}^{\infty}\frac{n-2}{n^{3}+2}

    (the same series, where the leading term has been removed). Since we know that removing a finite number of terms from a series will not affect convergence, this new series will converge if and only if the one in question converges. We notice that

    0\le\frac{n-2}{n^{3}+2}<\frac{n}{n^{3}+2}<\frac{n}{n^{3}}=\frac{1}{n^{2}}.

    (The reason we rewrite this series starting as 2 is because all of the terms need to be positive after a certain point in order to apply the comparison test… 2 is that point.) Since we know that the series

    \sum _{{n=2}}^{\infty}\frac{1}{n^{2}}

    converges, so does the series in question.

  3. Use the comparison test to determine if the given series converges or diverges.

    \sum _{{n=1}}^{\infty}\frac{5}{2n^{2}+4n+3}

    Solution: Notice that

    \frac{5}{2n^{2}+4n+3}<\frac{5}{2n^{2}}

    and

    \sum _{{n=1}}^{\infty}\frac{5}{2n^{2}}=\frac{5}{2}\sum _{{n=1}}^{\infty}\frac{1}{n^{2}}

    where we know the series on the right converges. Thus, the series in question converges.

  4. Use the comparison test to determine if the given series converges or diverges.

    \sum _{{n=1}}^{\infty}\frac{\ln n}{n}

    Solution: We used this example for the integral test as well. For the comparison test, we notice that when n>e we have

    \frac{\ln n}{n}>\frac{1}{n}

    and we know that the p-series

    \sum _{{n=1}}^{\infty}\frac{1}{n}

    diverges. Therefore, it follows from the comparison test that the given series diverges.

© 2011 Jason B. Hill. All Rights Reserved.