Why does the harmonic series diverge?

Posted on April 26, 2010 by Mr. Chase

My Precalculus students have been asking me this question. I don’t really have a great answer, except that it’s true. Granted it’s not very intuitive. But nothing about infinite series is intuitive. For those not in my class or not familiar with the harmonic series, the question is:

Does $\sum_{n=1}^{\infty}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$ converge?

And the answer is no. This is surprising, because the terms of the series approach zero.

The proof that it diverges is due to Nicole Oresme and is fairly simple. It can be found here. There are at least 20 proofs of the fact, according to this article by Kifowit and Stamps.

Interestingly, the alternating harmonic series does converge:

$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots =\ln{2}$

And so does the p-harmonic series with p>1. For instance:

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots =\frac{\pi^2}{6}$

Besides looking at the sequence of partial sums, I’m not sure I can help you with the intuition on any of these facts. Can you?

Really Fun Limit Problem (revisited)

Posted on April 20, 2010 by Mr. Chase

I posted this problem a few weeks ago:

The figure shows a fixed circle $C_1$ with equation $\left(x-1\right)^2+y^2=1$ and a shrinking circle $C_2$ with radius $r$ and center the origin (in red). $P$ is the point $(0,r)$ , $Q$ is the upper point of intersection of the two circles, and $R$ is the point of intersection of the line $PQ$ and the $x$ -axis. What happens to $R$ as $C_2$ shrinks, that is, as $r\rightarrow 0^{+}$ ?

And I also posted this applet, so you could investigate it yourself.

Now, I post the solution. Initially, I thought that the $x$ -coordinate of $R$ would go off to infinity or zero, I wasn’t sure which. The equation for $C_2$ is $x^2+y^2=r^2$ . Solving for the intersection of the two circles, $Q$ , we find it has coordinates

$Q=\left(\frac{r^2}{2},\sqrt{r^2-\frac{r^4}{4}}\right)$

Remembering that $P=(0,r)$ , we now find the equation of line $PQ$ in point-slope form.

$y-r=\left(\frac{r-\sqrt{r^2-\frac{r^4}{4}}}{0-\frac{r^2}{2}}\right)(x-0)$

Now, we seek to find the coordinates of $R$ , the $x$ -intercept of the line. Letting $y=0$ in the above equation, we solve for $x$ and find

$x=\frac{\frac{r^2}{2}}{1-\sqrt{1-\frac{r^2}{4}}}$

We now take the limit of $x$ as $r\rightarrow 0$ .

$\displaystyle\lim_{r\rightarrow 0}\frac{\frac{r^2}{2}}{1-\sqrt{1-\frac{r^2}{4}}}$

If we try to evaluate this limit by plugging in 0, we get an indeterminant form 0/0. We can either use L’hopital’s Rule or evaluate it numerically. Either way, we find.

$\displaystyle\lim_{r\rightarrow 0}x=4$

I have to say, this result surprised me. Like I said, I expected this limit to evaluate to 0 or infinity–but 4?? I had to get a bit more of an intuitive understanding, so I built that Geogebra applet.

Also, I was told afterward, by the student who brought this to me, that there’s a straight-forward geometric way of deriving the answer, based on the observation that point Q and point R and the point (2,0) (0,2) form an isosceles triangle for all values of r. The observation isn’t trivial. Can you prove it’s true? Once you do realize this fact, though, the above result is clear.

Really Interesting Limit Problem (again)

Posted on March 16, 2010 by Mr. Chase

I just posted this problem the other day.

I made this java applet to help visualize the problem. Give it a try. Use the slider to let r go to zero and see where point R goes. You’re sure to figure it out just by playing with the diagram. Now, can you prove it?