Day 40 in Mat229

• Our quiz covered the divergence and integral tests.
  • The Quiz
  • The Key
  • Median: 7.25
  • A few comments:
    1. Don't evaluate (unless you have to)! You're wasting time. Compare, and use theorems....
    2. The remainder term is \[ \sum_{n=N+1}^\infty \frac{1}{n^2} \] The integral for the upper bound starts at \(N\), rather than \(N+1\).

    3. The divergence test applies if the limit of the terms either doesn't exist, or exists but is non-zero.

       In either event, the series won't converge.

• I've assigned some problems from these two final chapters:
  • Try problems 250-257, 276-279, 280-281, 301-304 from Vol 2, Sec 5.5: Alternating Series
  • Try problems 317-323, 328-331, 338-341, 348-351 from Vol 2, Sec 5.6: Ratio and Root Tests.

• Please read Vol 2, Sec 6.1: Power Series and Functions for next time. This leads into the material we discussed during our lab on Friday:

  A major reason for all of this discussion of series of numbers is because we're going to be creating series to approximate functions (like sine); but then using partial sums (often of polynomial terms) to do the actual approximation.

Let's quickly review the AST, and the associated remainder estimate.

1. Alternating Series (.nb)
2. Alternating Series (.pdf)

• Definitions:

  • Absolutely convergent: A series is called absolutely convergent if the series of absolute values $\sum |a_n|$ converges.

  • Conditionally convergent: A series is called conditionally convergent if the series is convergent but not absolutely convergent.

• One thing that I can now mention is about re-arranging the terms of a series.
  • If the series is absolutely convergent, you can re-arrange terms willy-nilly: e.g. the series \[ \sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+1}\right) = \sum_{k=1}^\infty\frac{1}{k(k+1)} \]

    Note that if a positive-termed sequence is convergent, then it is absolutely convergent.

    So we can split this absolutely convergent series into even and odd terms, for example: \[ \sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+1}\right) = \sum_{k=1,odd}\left(\frac{1}{k}-\frac{1}{k+1}\right) + \sum_{k=1,even}\left(\frac{1}{k}-\frac{1}{k+1}\right) \] But notice that we cannot write \[ \sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+1}\right) = \sum_{k=1}^\infty\frac{1}{k} - \sum_{k=1}^\infty\frac{1}{k+1} \] We cannot break up the terms $a_k$ themselves and then re-arrange those (both series on the right are divergent -- harmonic).

  • If the series is only conditionally convergent, you cannot re-order the terms: e.g. \[ \sum_{k=1}^\infty(-1)^{k+1}\frac{1}{k} \] We cannot write the conditionally convergent series as \[ \sum_{k=1}^\infty(-1)^{k+1}\frac{1}{k} = \sum_{k=1,even}^\infty\frac{1}{k} - \sum_{k=1,odd}^\infty\frac{1}{k} \] Both the series on the right are divergent.

    Perhaps more astonishing, if a series is conditionally convergent you can rearrange the terms so that it will converge to anything! Let's have a look at the argument in the text, concerning the alternating harmonic series -- the poster child of conditionally convergent series.

  As an example of how rearrangement of terms is permitted for an absolutely convergent series, consider the problem of expressing $.\overline{467}$ as a ratio of integers:

  ---

  \[ .467467... = \frac{4}{10}+\frac{6}{100}+\frac{7}{1000}+ \frac{4}{10000}+\frac{6}{100000}+\frac{7}{1000000}+ \ldots \] We group terms in threes to produce \[ .467467... = \frac{467}{1000}+\frac{467}{1000^2}+ \ldots \] from which we obtain \[ .467467... = \frac{467}{1000}\sum_{k=0}^{\infty}\left(\frac{1}{1000}\right)^k = \frac{467}{1000}\frac{1}{1-\frac{1}{1000}} = \frac{467}{999} \] Alternatively, we can think of \[ .467467... = .400400... + .0600600... + .007007... \] (completely re-ordering the terms; but this is an absolutely convergent series, and so) \[ .467467... = \frac{400}{1000}\sum_{k=0}^{\infty}\left(\frac{1}{1000}\right)^k + \frac{60}{1000}\sum_{k=0}^{\infty}\left(\frac{1}{1000}\right)^k + \frac{7}{1000}\sum_{k=0}^{\infty}\left(\frac{1}{1000}\right)^k \] or \[ .467467... = \frac{400}{999} + \frac{60}{999} + \frac{7}{999} = \frac{467}{999} \]

  Exactly the same thing. Rearranging the terms didn't change anything.

  ---

• There are two more tests for convergence of series that we need to discuss:

  1. Ratio test: Let $\{a_k\}$ be a sequence and assume that the following limit exists: \[ r=\lim_{k\to \infty}\left|\frac{a_{k+1}}{a_k}\right| \]
     1. if $r<1$, then the series $\sum a_k$ converges absolutely.
     2. if $r>1$, then the series $\sum a_k$ diverges.
     3. if $r=1$, the test is inconclusive.
     This result says that eventually the ratio of successive terms is effectively constant, r: the terms of the sequence approach a "common ratio" as $k\to\infty$. What kind of series looks like that? A geometric series:

     \[ \sum_{k=0}^{\infty}{c}r^k \]

     The ratio test is effectively a self-referrential comparison test: we compare terms of $a_n$ with other terms $a_{n+1}$ (rather than with some other series).

  2. Root test: Let $\{a_k\}$ be a sequence and assume that the following limit exists: \[ r=\lim_{k\to \infty}\sqrt[k]{|a_k|} \]
     1. if $r<1$, then the series $\sum a_k$ converges absolutely.
     2. if $r>1$, then the series $\sum a_k$ diverges.
     3. if $r=1$, the test is inconclusive.

     This result says that eventually the absolute values of the terms are effectively equal to $|a_k|={r^k}$: what kind of series looks like that? A geometric series!

     Notice that, once again, limits of sequences plays an important role. Series are just sums of sequences, after all; we're focused on how terms behave (root test), how successive terms behave (ratio test), or how partial sums behave.

1. (Absolute.nb)
2. (Absolute.pdf)