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Having fun teaching Σ of geometric series, and a π question for you.

March 17, 2010 am31 7:04 am

I don’t know why I am teaching sums of geometric sequences in algebra 2. Actually, I know, but the reason makes no sense.

  • In any case, I’m teaching them. And having fun. Teaching is better (for them, for me) when I’m having fun. When they’re having fun. When the learning feels like play.

    Anyway, they now know that if the ratio between successive terms is between -1 and 1, then we can find the sum.  For example:  10 – 1 + .1 – .01 + … = \frac{10}{1 - -0.1} = \frac{100}{11}. Cool.

    Actually, I liked how I started. “Draw a square. Give it an area. Raise a rectangle on the side of the square, with half the area of the square. Note the total area. Raise a new rectangle, half the area of the previous. Note the total area. Keep going until your composite figure is double the area of your original square.” Immediately some kids started squealing that it wasn’t possible, but I would have none of it. “You started with 16, and you are already up to 31.5 – almost there. Keep going.”  In another minute the squeaks turned into a chorus, and I confessed.

    Then I drew a 2:1 rectangle, and filled in half its emptiness. Repeat, repeat, repeat, to general agreement that we could fill in as much as we like, but we’d never be done.

    Fast forward. I put up \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+ ... They agree its not geometric, and I ask them to look for the sum anyhow. “It’s slowing down, but maybe it doesn’t stop…” “Keep going. Notice anything special happening at π?” Of course they didn’t as they shot past it. But it kept them focused a bit longer. “No, it doesn’t look like it’s reaching a number” I egged them towards 2π, but gave them a break before they found that it runs right past that number as well. They have precalc to investigate this more fully.

    But since they were playing along so nicely, I offered them this:  \frac{1}{1}-\frac{1}{2}+\frac{1}{3}- \frac{1}{4}+\frac{1}{5}- .... I made them rewrite it in sigma notation (always a cool exercise), and then asked them to play. We graphed their very partial results. They know it is point six something. But I don’t remember. Is the sum 2/π?  Why?  And how long does it take to become clear?

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    10 Comments leave one →
    1. Sara permalink
      March 17, 2010 am31 7:54 am 7:54 am

      You can apply the alternating series test (limit of each term is 0, monotone decreasing w/ alternating signs) to show that it converges. The value is ln 2. I bet if you google “Alternating Harmonic Series” you can find a proof.

    2. March 17, 2010 am31 11:15 am 11:15 am

      JD,
      I like to make sure they get the idea that the summation of 1/2^n =1 is not something that is just approached, or un-doable because it goes on for ever… with an example that is already done…. Make a really long line segment and say, that’s one… they will agree.. now ask them if there is a point that is 1/2 unit away from one end..a midpoint… and a point half way between there and the right end? and half way between there and the right end again??? and of course the points, and the infinite number of line segments are already there,,, and they all add up to one.. Not infinity in the making…. infinity, DONE.

    3. Natasha permalink
      March 17, 2010 pm31 1:53 pm 1:53 pm

      It is convergent (alternating series test, as noted previously) and the sum is ln(2) (I used Maple); showing this analytically takes a few steps. I’m going to TeXify my notation for convenience. (I wonder if that’ll send me to the spam filter, as it looks nonsense. I’ll email a PDF as well so that it’s readable.)

      Source: Stewart’s Calculus, 5th ed.

      Consider the series, written $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}$, which is alternating harmonic. Additionally consider the corresponding harmonic series $\sum_{n=1}^\infty \frac{1}{n}$, which we know diverges. Relate the partial sums of both series by
      \[s_{2n} = h_{2n} – h_n,\]
      where $s_n$ is the $n$th partial sum of the series at hand and $h_n$ the $n$th partial sum of the corresponding harmonic series.

      Use the fact that the limit of $h_n – \ln(n)$ as $n$ approaches infinity is Euler’s constant, $\gamma$. (This fact takes a couple of steps to show, but it’s not any harder than the rest of this problem.) Then $\lim\limits_{n\to \infty} h_{2n} – \ln(2n) = \gamma$.

      Use with the relation between $s_{2n}$, $h_n$ and $h_{2n}$ to show that $s_{2n} \rightarrow \ln(2)$ as $n \rightarrow \infty$:

      First, we have
      \begin{align*}
      \lim\limits_{n\to\infty} \left(h_{2n}-\ln(2n)\right) &=\gamma \\
      \lim\limits_{n\to\infty} \left(h_{2n}-(\ln(2)+\ln(n))\right) &=\gamma \\
      \lim\limits_{n\to\infty} \left(h_{2n}-\ln(n)\right) &=\gamma + \ln(2) \\
      \end{align*}
      Also, $\lim\limits_{n\to\infty}\left(h_n-\ln(n)\right) = \gamma$. Therefore
      \begin{align*}
      \lim\limits_{n\to\infty} s_{2n} &= \lim\limits_{n\to\infty} \left(h_{2n} -h_n \right) \\
      \Leftrightarrow \quad \lim\limits_{n\to\infty} s_{2n} &= \lim\limits_{n\to\infty} \left(h_{2n} – \ln(n) – (h_n – \ln(n)) \right) \\
      \Leftrightarrow \quad \lim\limits_{n\to\infty} s_{2n} &= \lim\limits_{n\to\infty} \left(h_{2n} – \ln(n)\right) – \lim\limits_{n\to\infty}\left(h_n – \ln(n) \right) \\
      \Leftrightarrow \quad \lim\limits_{n\to\infty} s_{2n} &= \gamma + \ln(2) – \gamma \\
      \Leftrightarrow \quad \lim\limits_{n\to\infty} s_{2n} &= \ln(2).
      \end{align*}

      Whew! That looks nasty in a comment box. Hope my mathematics is sound.

    4. Natasha permalink
      March 17, 2010 pm31 2:03 pm 2:03 pm

      also, whoops… re: the limit of the difference between the harmonic series and ln(n) as n approaches infinity… Showing existence is easy. Showing that it equals Euler’s constant, well, Euler’s constant is defined as that difference. Pardon me.

    5. March 17, 2010 pm31 6:40 pm 6:40 pm

      This is Natasha’s comment, formatted:
      (Natasha, I don’t know if your comment was left unformatted on purpose. In wordpress, dollar_sign followed by the word latex (no space) begins the text, and a regular dollar sign ends it)

      It is convergent (alternating series test, as noted previously) and the sum is ln(2) (I used Maple); showing this analytically takes a few steps. I’m going to TeXify my notation for convenience. (I wonder if that’ll send me to the spam filter, as it looks nonsense. I’ll email a PDF as well so that it’s readable.)

      Source: Stewart’s Calculus, 5th ed.

      Consider the series, written \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}, which is alternating harmonic. Additionally consider the corresponding harmonic series \sum_{n=1}^\infty \frac{1}{n}, which we know diverges. Relate the partial sums of both series by
      [s_{2n} = h_{2n} - h_n,]
      where s_n is the nth partial sum of the series at hand and h_n the nth partial sum of the corresponding harmonic series.

      Use the fact that the limit of h_n - \ln(n) as n approaches infinity is Euler’s constant, \gamma. (This fact takes a couple of steps to show, but it’s not any harder than the rest of this problem.) Then \lim\limits_{n\to \infty} h_{2n} - \ln(2n) = \gamma.

      Use with the relation between s_{2n}, h_n and h_{2n} to show that s_{2n} \rightarrow \ln(2) as n \rightarrow \infty:

      First, we have
      \begin{align*}
      \lim\limits_{n\to\infty} \left(h_{2n}-\ln(2n)\right) &=\gamma \\
      \lim\limits_{n\to\infty} \left(h_{2n}-(\ln(2)+\ln(n))\right) &=\gamma \\
      \lim\limits_{n\to\infty} \left(h_{2n}-\ln(n)\right) &=\gamma + \ln(2) \\
      \end{align*}
      Also, $\lim\limits_{n\to\infty}\left(h_n-\ln(n)\right) = \gamma$. Therefore
      \begin{align*}
      \lim\limits_{n\to\infty} s_{2n} &= \lim\limits_{n\to\infty} \left(h_{2n} -h_n \right) \\
      \Leftrightarrow \quad \lim\limits_{n\to\infty} s_{2n} &= \lim\limits_{n\to\infty} \left(h_{2n} – \ln(n) – (h_n – \ln(n)) \right) \\
      \Leftrightarrow \quad \lim\limits_{n\to\infty} s_{2n} &= \lim\limits_{n\to\infty} \left(h_{2n} – \ln(n)\right) – \lim\limits_{n\to\infty}\left(h_n – \ln(n) \right) \\
      \Leftrightarrow \quad \lim\limits_{n\to\infty} s_{2n} &= \gamma + \ln(2) – \gamma \\
      \Leftrightarrow \quad \lim\limits_{n\to\infty} s_{2n} &= \ln(2).
      \end{align*}

      Whew! That looks nasty in a comment box. Hope my mathematics is sound.

    6. DavidC permalink
      March 18, 2010 am31 1:29 am 1:29 am

      Glad you’re having fun!

      Seems like series are something students could (did, in this case!) have a lot of fun playing with. And since it’s Algebra II, you probably aren’t going to kill the subject with a battery of convergence tests they have to memorize and apply…

      Wikipedia points out that the alternating harmonic series is what you get from evaluating the Taylor series for \ln(1+x) at 1. Cool!

      (Not that that will help your students see why it’s $\latex ln(2)$, I guess. If your students even know about \ln and exp? But I’m not sure why they would know about those? Are those functions taught in algebra? Can the number e make sense and be well motivated before calculus?)

      Since you asked “And how long does it take to become clear?”: I think Taylor’s Theorem should provide bounds on the rate of convergence. I think I’m finding (I used the Lagrange form of the theorem) that if you use a sum with n terms after the ‘1’, then the size of the error will be between \frac{1}{2^n} and \frac{1}{n}. Maybe someone more clever than me can give you tighter bounds, or see that I’m wrong.

    7. DavidC permalink
      March 18, 2010 am31 1:48 am 1:48 am

      Ack, I think I am not quite right about the bounds. Sorry for not explaining my reasoning, but typing tex is hard when it’s bedtime.

    8. March 21, 2010 am31 12:34 am 12:34 am

      The similar series relevant to pi (maybe what you were thinking of?) is 1 – 1/3 + 1/5 – 1/7 + … which converges to pi/4 (the “Leibniz series“), but very slowly.

      • March 21, 2010 am31 8:27 am 8:27 am

        Yes Ben, thanks, that was probably in the back of my mind.

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