Integral Test

Section: 
9.3
Date: 
Monday, October 4, 2010 - 14:00 - 15:00
AttachmentSize
fall2010math2300_integral-test.pdf58.78 KB

Math 2300 Section 005 – Calculus II – Fall 2010

9.3 Integral Test Examples – October 4, 2010

  1. What does Theorem 9.2 say about the convergence or divergence of the series?

    \sum _{{n=1}}^{\infty}\left(2+e^{{-n}}\right)

    Solution: Since the series terms are of the form 2+e^{{-n}}, and the limit of these terms as n\rightarrow\infty is 2 (not zero), we know that this series diverges.

  2. What does Theorem 9.2 say about the convergence or divergence of the series?

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

    Solution: Again, since the series terms do not converge to zero (in this case they don't converge at all), the series diverges.

  3. What does Theorem 9.2 say about the convergence or divergence of the series?

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

    Solution: Theorem 9.2 says nothing in this situation. Remember that part (3) of the theorem states that when the series terms do not converge to zero then the series itself will not converge… but it says nothing about what happens when the series terms go to zero. In fact, we may create many series where the terms approach zero and the series itself is divergent.

  4. Harmonic Series is Divergent: Show that the “Harmonic Series,”

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

    diverges.

    Solution: See the graph for Example 2 on page 478. We notice that the partial sums S_{n} satisfy the relation

    S_{n}=\sum _{{i=1}}^{n}\frac{1}{i}>\int _{1}^{{n+1}}\frac{1}{x}\, dx.

    (That is, the rectangles create an area that is larger than the area under the curve y=1/x.) Thus, we have

    \lim _{{n\rightarrow\infty}}S_{n}>\lim _{{n\rightarrow\infty}}\int _{1}^{{n+1}}\frac{1}{x}\, dx

    where we know that the improper integral on the right diverges. Therefore, the harmonic series diverges.

  5. Likewise, show that the series

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

    converges.

    Solution: Look at the graph in example 3 on page 478. We use Theorem 9.2 after rewriting the series as

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

    If the series on the right converges, then so will the series on the left. We have

    S_{n}-1=\sum _{{i=2}}^{n}\frac{1}{i^{2}}<\int _{1}^{{n+1}}\frac{1}{x^{2}}\, dx

    and taking the limit as n\rightarrow\infty gives us convergence.

  6. Use the integral test to determine if the series converges or diverges.

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

    Solution:

    \displaystyle\int _{1}^{\infty}\frac{1}{x^{2}+1}\, dx \displaystyle=\lim _{{b\rightarrow\infty}}\int _{1}^{b}\frac{1}{x^{2}+1}\, dx
    \displaystyle=\lim _{{b\rightarrow\infty}}\left[\tan^{{-1}}(x)\right]_{1}^{b}
    \displaystyle=\lim _{{b\rightarrow\infty}}\tan^{{-1}}b-\frac{\pi}{4}
    \displaystyle=\frac{\pi}{2}-\frac{\pi}{4}\,=\,\frac{\pi}{4}

    Thus, by the integral test, the series converges.

  7. Use the integral test to determine if the series converges or diverges.

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

    Solution: This is a bit more challenging, since the function is not always decreasing. In order to use the integral test, we need to know that the function is decreasing and positive past some point… since removing the values of the function before that point (to the left on the x-axis) will not affect convergence by Theorem 9.2 (2). We find from the derivative that the function is always decreasing (and positive) when n>e. Thus, we're really interested in the convergence of

    \displaystyle\int _{e}^{\infty}\frac{\ln x}{x}\, dx \displaystyle=\lim _{{b\rightarrow\infty}}\int _{e}^{b}\frac{\ln x}{x}\, dx
    \displaystyle=\lim _{{b\rightarrow\infty}}\int _{e}^{b}u\, du
    \displaystyle=\lim _{{b\rightarrow\infty}}\left[\frac{u^{2}}{2}\right]_{e}^{b}
    \displaystyle=\lim _{{b\rightarrow\infty}}\left[\frac{(\ln x)^{2}}{2}\right]_{e}^{b}
    \displaystyle=\lim _{{b\rightarrow\infty}}\frac{(\ln b)^{2}}{2}-\frac{1}{2}
    \displaystyle=\infty

    By the integral test, the series diverges.

  8. Use the integral test to determine if the series converges or diverges.

    \sum _{{n=1}}^{\infty}ne^{{-n}}

    Solution: We need to use integration by parts on an improper integral.

    \displaystyle\int _{1}^{\infty}xe^{{-x}}\, dx \displaystyle=\lim _{{b\rightarrow\infty}}\int _{1}^{b}xe^{{-x}}\, dx
    \displaystyle=\lim _{{b\rightarrow\infty}}\left[\left[-xe^{{-x}}\right]_{1}^{b}+\int _{1}^{b}e^{{-x}}\, dx\right]
    \displaystyle=\lim _{{b\rightarrow\infty}}\left[-be^{{-b}}+\frac{1}{e}-\int _{{-1}}^{{-b}}e^{x}\, dx\right]
    \displaystyle=\lim _{{b\rightarrow\infty}}\left[-be^{{-b}}+\frac{1}{e}-\left[e^{{-b}}-\frac{1}{e}\right]\right]
    \displaystyle=\frac{2}{e}

    Thus, the series converges.

© 2011 Jason B. Hill. All Rights Reserved.