What follows below reflects uncategorized recent activity in the courses I am teaching, with the most recent activity at the top. This information can be categorized by course and type by selecting the "courses" link above.

Ratio Test

Section: 
9.4
Date: 
Wednesday, October 6, 2010 - 14:00 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

9.4 Ratio Test Examples – October 6, 2010

  1. What does the ratio test say about the series?

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

    Solution:

    \lim _{{n\rightarrow\infty}}\frac{\displaystyle\frac{1}{n+1}}{\displaystyle\frac{1}{n}}=\lim _{{n\rightarrow\infty}}\frac{n}{n+1}=1

    and so the ratio test says nothing about the series. Of course, we already know that the harmonic series is divergent.

Comparison Test

Section: 
9.4
Date: 
Tuesday, October 5, 2010 - 14:00 - Wednesday, October 6, 2010 - 15:00

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.

Integral Test

Section: 
9.3
Date: 
Monday, October 4, 2010 - 14:00 - 15:00

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

Exam 2 Review 1

A review sheet for exam 2 is now available, as an attachment to this post.

LA Information

Date: 
Wednesday, October 6, 2010 - 18:00 - 19:00

There is an informational session about becoming a Learning Assistant (LA) on Wednesday, October 6 at 6PM in UMC235. Applications for Spring 2011 will be available October 6 - October 20.

If you are interested in attending the informational session, e-mail olivia.holzman@colorado.edu

Fibonacci Sequence Video

Someone pointed this out to me on YouTube. It's a video about the Fibonacci Sequence.

Geometric Series

Section: 
9.2
Date: 
Friday, October 1, 2010 - 14:00 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

9.2 Geometric Series Notes – October 1, 2010

A tip on recognizing geometric series: Many times when you're using convergence comparison tests, the series you use for comparison is either a p-series or a geometric series. In the first case, we know that

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

converges for p>1 and diverges for p\le 1. In the second case, the geometric series in question may be more challenging to pick out. For instance, when you have

\sum _{{n=1}}^{\infty}\frac{1}{2^{n}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots

you may either view this as the geometric series

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

where a=1/2 and x=1/2, giving from the formulas for geometric series that

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

Or (and this may seem odd at first, but I think it yields faster computations) view this as the geometric series

1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots

where a=1 and x=1/2, except that the first term needs to be removed (since the series in question has no 1). So, we also have

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

Webwork 9.2 Hints

Math 2300 Section 005 – Calculus II – Fall 2010

Hints for Webwork Section 9.2

Most of the Webwork problems for this section simply use the notions of geometric series that we developed in class. There are a few, however, that are challenging and require you to put a lot of thought in to setting up the series.

5. Consider a similar example. Us math folk really like coffee. Assume that at the start of each day I drink 300mg of caffeine. Also assume that the “halflife” of this caffeine in my system is 7.5 hours. So, in 7.5 hours I will effectively have 150mg of caffeine in my system, and in 15 hours that will be halved again and I will now have 75mg of caffeine in my system. Notice that at 24 hours (a full day after my last sip of coffee) I still have some caffeine in my system. How much? The answer: I take my initial amount and multiply it by half for each period of 7.5 hours. And since 24/7.5 of those periods happen in 24 hours, we find that I have

300\cdot\frac{1}{2}^{{24/7.5}}\approx 32.64\,\text{mg}

of caffeine in my system after 24 hours. At that instant, I drink another 300mg of caffeine and I now have

Sequence and Series Quiz

Due Date: 
Friday, October 1, 2010 - 16:00 - Monday, October 4, 2010 - 16:00

This quiz is due on Monday at 4PM. Obviously, only questions 1 and 2 count for credit. Due to my current lack of creativity, there is no replacement question on this quiz.

Worksheet 5

The solutions are posted to worksheet #5

© 2011 Jason B. Hill. All Rights Reserved.