Having fun teaching Σ of geometric series, and a π question for you.
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 + … = 
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 
But since they were playing along so nicely, I offered them this: 
JD2718 
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.
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.
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.
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.
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
, which is alternating harmonic. Additionally consider the corresponding harmonic series 
, which we know diverges. Relate the partial sums of both series by
![[s_{2n} = h_{2n} - h_n,]](https://s0.wp.com/latex.php?latex=%5Bs_%7B2n%7D+%3D+h_%7B2n%7D+-+h_n%2C%5D&bg=ffffff&fg=333333&s=0&c=20201002)
is the 
th partial sum of the series at hand and 
the 
th partial sum of the corresponding harmonic series.
where
Use the fact that the limit of
as 
approaches infinity is Euler’s constant, 
. (This fact takes a couple of steps to show, but it’s not any harder than the rest of this problem.) Then 
.
Use with the relation between
, 
and 
to show that 
as 
:
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.
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
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
and exp? But I’m not sure why they would know about those? Are those functions taught in algebra? Can the number 
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
terms after the ‘1’, then the size of the error will be between 
and 
. Maybe someone more clever than me can give you tighter bounds, or see that I’m wrong.
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.
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.
Yes Ben, thanks, that was probably in the back of my mind.