Midterm 2 Review Quiz

Due Date: 
Tuesday, October 12, 2010 - 16:00
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Math 2300 Section 005 – Calculus II – Fall 2010

Quiz – Due Tuesday, October 12, 2010

This quiz is intended to be a review, so it contains more problems than usual. I won't be grading each problem individually, but the quiz will be graded as a whole. Solutions will be posted on Tuesday.

  1. Find the sum

    \sum _{{n=3}}^{{12}}\frac{3^{n}+6}{4^{n}}

    Hint: Write this as two geometric series.

    \displaystyle\sum _{{n=3}}^{{12}}\frac{3^{n}+6}{4^{n}} \displaystyle=\sum _{{n=3}}^{{12}}\left(\frac{3^{n}}{4^{n}}+\frac{6}{4^{n}}\right)
    \displaystyle=\sum _{{n=3}}^{{12}}\frac{3^{n}}{4^{n}}+\sum _{{n=3}}^{{12}}\frac{6}{4^{n}}
    \displaystyle=\sum _{{n=3}}^{{12}}\left(\frac{3}{4}\right)^{n}+\sum _{{n=3}}^{{12}}6\left(\frac{1}{4}\right)^{n}

    You should be able to solve from here.

  2. Do the following sums converge or diverge?

    1. \displaystyle\sum _{{n=0}}^{\infty}ne^{{-n^{2}}}

      Hint: The first thing I think when I see this is that the integral

      \int _{0}^{\infty}xe^{{-x^{2}}}\, dx

      can be done using a substitution. I also notice that the function f(x)=xe^{{-x^{2}}} is positive. The derivative f^{{\prime}}(x) is given by

      f^{{\prime}}(x)=e^{{-x^{2}}}-2x^{2}e^{{-x^{2}}}=(1-2x^{2})e^{{-x^{2}}}

      and setting this equal to zero shows that we have critical points where x=\pm\sqrt{1/2}. You may want to convince yourself that this function is indeed decreasing beyond, say, x=1. Then, consider convergence of the series

      \sum _{{n=1}}^{\infty}ne^{{-n^{2}}}
    2. \displaystyle\sum _{{n=1}}^{\infty}\frac{n!(n+1)!}{(2n)!}

      Hint: The obvious way to approach series with factorials is with the ratio test. I'll run through the limit calculation here and let you decide the result. You should know how to calculate with factorials.

      \displaystyle\lim _{{n\rightarrow\infty}}\frac{\displaystyle\left|\frac{(n+1)!(n+2)!}{(2n+2)!}\right|}{\displaystyle\left|\frac{n!(n+1)!}{(2n)!}\right|} \displaystyle=\lim _{{n\rightarrow\infty}}\frac{\displaystyle\frac{(n+1)!(n+2)!}{(2n+2)!}}{\displaystyle\frac{n!(n+1)!}{(2n)!}}
      \displaystyle=\lim _{{n\rightarrow\infty}}\frac{\displaystyle(n+1)!(n+2)!(2n)!}{\displaystyle(2n+2)!n!(n+1)!}
      \displaystyle=\lim _{{n\rightarrow\infty}}\frac{\displaystyle(n+2)!(2n)!}{\displaystyle(2n+2)!n!}
      \displaystyle=\lim _{{n\rightarrow\infty}}\frac{(n+2)(n+1)}{(2n+2)(2n+1)}
      \displaystyle=\lim _{{n\rightarrow\infty}}\frac{n^{2}+3n+2}{4n^{2}+6n+3}

      It's up to you to do the calculation from here.

    3. \displaystyle\sum _{{n=2}}^{\infty}\frac{(-1)^{{n-1}}}{\sqrt{n^{2}+1}}

      Hint: Generally, the first thing you want to do when you see an alternating series is to use the alternating series test. This series is clearly alternating and satisfies the property needed in the test. Can you explain why? What is the result of the test?

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

      Hint: This one is a bit more random. First, since

      \frac{n^{2}}{n^{3}+1}<\frac{n^{2}}{n^{3}}=\frac{1}{n}

      shows that our function is bounded above by a divergent function, we don't get anything immediately from the comparison test. Maybe we can use the limit comparison test instead. Let

      a_{n}=\frac{n^{2}}{n^{3}+1}\qquad\text{and}\qquad b_{n}=\frac{1}{n}.

      Then we find that

      \displaystyle\lim _{{n\rightarrow\infty}}\frac{a_{n}}{n_{n}} \displaystyle=\lim _{{n\rightarrow\infty}}\frac{\displaystyle\frac{n^{2}}{n^{3}+1}}{\displaystyle\frac{1}{n}}
      \displaystyle=\lim _{{n\rightarrow\infty}}\frac{n^{3}}{n^{3}+1}

      and I'll let you finish from here.

    5. \displaystyle\sum _{{n=1}}^{\infty}\ln\left(1+\frac{1}{n}\right)

      Hint: Notice that

      \ln\left(1+\frac{1}{n}\right)=\ln\left(\frac{n}{n}+\frac{1}{n}\right)=\ln\left(\frac{n+1}{n}\right)=\ln(n+1)-\ln(n)

      and while it is clear that the derivative is negative (i.e., for n>e the differences will be less and less) we find that

      \frac{d}{dn}\left(\ln(n+1)-\ln(n)\right)=\frac{1}{n+1}-\frac{1}{n}.

      This is a tricky problem, but what you may want to try to do now is to relate this function to a p-series that we know diverges. I'll let you try to finish the next portion, or find a better way of doing the problem in general.

  3. Find the radius of convergence of

    \sum _{{n=1}}^{\infty}\frac{(2n)!x^{n}}{(n!)^{2}}

    This won't be on the exam. Try to complete it, but don't put it at such a priority as the other problems.

  4. Find the interval of convergence of

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

    This won't be on the exam. Try to complete it, but don't put it at such a priority as the other problems.

© 2011 Jason B. Hill. All Rights Reserved.