Day 36 in MAT229

Today:

• Announcements:
  • How was lunch?

  • Abby's being mis-treated by IMath on 11.3, problem 9. If you are too, please let me know. I think that I've got it fixed now. If you already attempted it, you'll have to attempt it again (it was probably mis-treating you, too!)

• Today:
  • I want to start by talking a little about approximating series with partial sums to a given tolerance.

    What's this all about?

    Suppose I have the series \[ \sum_{k=1}^\infty\frac{1}{2^k} = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \] The terms represent "room distances", and I start from the far wall.

    Suppose I have a sandwich, and want to hand it to someone outside in the hallway, and that I have to be able to reach out to give it to them (they can't stick their hand in).

    Then I have to get close enough for my arm to go through the door frame. How many terms will I need of the series to hand over the sandwich?

    1. My arm length is the tolerance;
    2. $n$ is the number of terms I'll need to go to get that close;
    3. and $S_n$ will be the distance I've traveled at the moment I hand over the sandwich.
    In this case, we can compute $S$ itself, of course ($S=1$ -- it's right to the threshold of the door); but we're usually talking about approximating more complex series (perhaps leveraging off a geometric or p-series, whose error we know how to bound).

    So $|S-S_n|<{\textrm{arm's length}}$

    means taking $n$ terms and getting to within range to hand over the sandwich.

  • Last time we wrapped up alternating series, and started Sect. 11.6: Absolute convergence test (pdf, pdf summary)

    One thing that I can now mention is about re-arranging the terms of a series. I've told you that you cannot do this willy-nilly:

    • 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.
    As an example of how to use this, I would like to reconsider the problem of expressing $.\overline{467}$ as a ratio of integers.

  • There are two more tests for convergence of series within 11.6 that we want 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 with other terms (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!

       No wonder the results of the tests look exactly the same.... Too bad we didn't just use the same letter for the limits in both cases!

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

       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.