Rolfe Schmidt

I’m learning. Slowly.

A Few More Series

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As I was sketching out my argument that \sum_{p \textbf{ prime} } \frac{1}{p} goes to infinity, I realized that there are a few other infinite sums that I need to discuss first. This is turning into a series of longish posts, so I want to make my goal here clear. I believe that any interested student with background in what americans call “Algebra II” can understand many problems and techniques that are usually reserved for university Math majors and grad students. Most of what I’m saying in these posts can be dispensed with in a few lines of Calculus, but I don’t want to use that crutch. I am writing to that interested student starting out, and trying to convince them that the big ideas are accessible to them now. You don’t need years of training to get to the good stuff, you should spend your years of training doing the good stuff.

Hopefully by writing posts like this and refining them, I’ll do a better job helping my own kids out when they are ready.

On that note, back to work. I’ll start by looking at another sum that goes to infinity, the harmonic series.

The harmonic series

If we want to get a good idea about how \sum_{p \textbf{ prime} } \frac{1}{p} might behave, we can look at the simpler and larger sum

\sum_{n = 1}^{\infty} \frac{1}{n}

This is called the harmonic series. This is just like \sum_{p \textbf{ prime} } \frac{1}{p}, except now we are summing over all whole numbers, not just primes. If “a lot of numbers are prime”, then the difference shouldn’t be too great. If “very few numbers are prime”, then the difference should be large. So measuring the difference between these two sums is really the same as understanding the distribution of the prime numbers. More on that later, now let’s get to work.

I claim that the harmonic series diverges. It will keep growing and growing. To see why, let’s break it into chunks. The first chunk will be pretty small: it is just the first term, 1. The second chunk has two terms: \frac{1}{2} + \frac{1}{3}. The next chunk has 4 terms: \frac{1}{4} + \frac{1}{5}  + \frac{1}{6} + \frac{1}{7}. The next chunk has 8 terms, then 16, and so on. Each chunk has twice as many terms as the one before it.

Now look at the size of the terms in each chunk. The first chunk has one term, and it has a value of 1. The second chunk has 2 terms — \frac{1}{2} and \frac{1}{3} and both of them are certainly bigger than \frac{1}{4}. So the sum of terms in this chunk has to be bigger than \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.

The next chunk has 4 terms and they’re all bigger than \frac{1}{8}, so the sum of terms in the next chunk must be bigger than 4 \times \frac{1}{8} = \frac{1}{2}.

The next chunk has 8 terms bigger than \frac{1}{16}. And so on. Here is a little picture of the first few terms. The red bars show the values \frac{1}{n} and the green boxes show the lower bounds.

Bound the Harmonic Series from below

The n-th chunk has 2^{n-1} terms, and they are all bigger than \frac{1}{2^{n-2} }. This means that when we add up the terms in each chunk we get some number bigger than one half. But there are an infinite number of chunks! If we add \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ... an infinite number of times, of course the sum will keep growing and growing. It diverges.

And the harmonic series is even bigger:

\sum_{n=1}^{\infty} \frac{1}{n} = \sum_{\textbf{ chunks} } \textbf{sum for chunk} > \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...

So we can safely say that \sum_{n=1}^{\infty} \frac{1}{n} = \infty. The sum diverges.

This process of breaking a sum into manageable chunks, each one twice as big as the last, is called a dyadic decomposition.

An almost-harmonic series

The numbers \frac{1}{n} get smaller and smaller, but they don’t get small fast enough to make the harmonic series converge. what if we made the terms smaller? Consider the series

\sum_{n=1}^{\infty}\frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ...

We can play a similar game with this series, breaking the sum into chunks. Only this time we are going to show that the sum of terms in each chunk is smaller than some small number.

The first chunk is easy, it has one term and the sum is just one. In the next chunk, notice that each term is smaller than \frac{1}{4}. There are two terms, so the sum for this chunk is less than 2 \times \frac{1}{4} = \frac{1}{2}.

For the third chunk, there are 4 terms, and each of them is smaller than \frac{1}{16} so the sum of terms in this chunk is less than 4 \times \frac{1}{16} = \frac{1}{4}. For the fourth chunk, we find the sum is smaller than 8 \times \frac{1}{64} = \frac{1}{8}, the sum of terms in the fith chunk is smaller than 16 \times \frac{1}{256} = \frac{1}{16} and so on.

Putting these together we find that

\sum_{n=1}^{\infty}\frac{1}{n^2} < 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... = \sum_{i=0}^{\infty} \frac{1}{2^i}

This is just the geometric series we saw in the last post, and int converges to \frac{1}{1 - \frac{1}{2} } = 2. So we know that

\sum_{n=1}^{\infty} \frac{1}{n^{2} } < 2

Not only does the sum converge, it is pretty small. The exact value of the sum was found by Leonhard Euler (it was called the Basel problem), and is \frac{\pi^{2}}{6} = 1.644934....

We can also look at series that lie between this one and the harmonic series. Consider any number s > 1. We can look at the series

\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}

A slight variation of the argument I just gave shows that this sum converges as long as s > 1. (Excercise: [1] Prove this for s = 1.1. [2] Prove this for all s > 1.) So we can talk sensibly about this sum \zeta(s), and study the way it changes as s changes.

\zeta is known as the Riemann zeta function. It is the function that lies at the heart of the Riemann Hypothesis — perhaps the most famous unsolved Math problem today. As we’ll start to see in the next post, if we can understand zeta, then we understand the primes.

Written by Rolfe Schmidt

March 12, 2008 at 9:53 am

Posted in Math

2 Responses

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  1. [...] make it meaningful. We weren’t interested in adding up whole numbers anyway. In our last post we found that when we can define the Riemann zeta [...]

  2. [...] are still my favorites, especially the ones collected here and the little series about the Riemann zeta function (which I really should [...]


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