# The harmonic series

The harmonic series is the expression

$1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \dotsb,$

whose value is an infinite sum. Since the terms in the series become arbitrarily small as one goes further in (since for every positive real number $\varepsilon$ and every positive integer $n$ greater than $\lceil \frac 1 \varepsilon \rceil$, the $n$th term is smaller than $\varepsilon$), one might expect the series to converge a finite value, but this is not the case—its value is actually infinite. Having terms which become arbitrarily small eventually is a necessary, but not sufficient, property of convergent series, and the harmonic series is the usual example brought up to show that this property is not sufficient.

Proving that the value of the harmonic series is infinite can be done easily, but not in an obvious manner. It involves grouping the terms in a clever manner so that the series can be seen to have the same value as one which obviously diverges to infinity. $\frac 1 2$ on its own makes up the first group. The next group contains $\frac 1 3$ and $\frac 1 4$. The next group contains 4 terms, $\frac 1 5$, $\frac 1 6$, $\frac 1 7$ and $\frac 1 8$. It continues on in the same way, with each group containing twice as many terms as the last group, so that the last terms in each group are the powers of two. The first term, 1, is outside of this sequence, so it can be considered not part of any group.

Now, consider the series obtained by replacing every term with the power of two that is the final one in its group. This series will look like this:

$1 + \frac 1 2 + \frac 1 4 + \frac 1 4 + \frac 1 8 + \frac 1 8 + \frac 1 8 + \frac 1 8 + \dotsb$

None of the terms have been increased in size by this replacement—they have only gotten smaller if they changed at all. Therefore, if we can show that the sum of this new series is infinite, the sum of the harmonic series, since it must be larger, must also be infinite. To see that the sum of the new series is infinite, just add up the terms in each group. The first group just contains $\frac 1 2$. The second group contains 2 instances of $\frac 1 4$; these add up to $\frac 1 2$. The third group contains 4 instances of $\frac 1 8$; these add up to $\frac 1 2$ as well. Clearly, every single group adds up to $\frac 1 2$, so the series is the same as the following one:

$1 + \frac 1 2 + \frac 1 2 + \frac 1 2 + \dotsb$

which is obviously infinite. This completes the proof that the value of the harmonic series is infinite.

A similar strategy can be used to examine all the series of the form

$1 + \frac 1 {2^p} + \frac 1 {3^p} + \frac 1 {4^p} + \dotsb$

where $p$ is an arbitrary real number. Clearly if $p < 1$, every term in this series will be greater than the corresponding term in the harmonic series, so its value must be infinite since it has to be larger than the value of the harmonic series. So the only interesting case is where $1 < p$.

If you try grouping the terms in the exact same way here, you’ll find that the first group will add up to $\frac 1 {2^p}$, the second will add up to $\frac 2 {4^p} = \frac 1 {2^{2 p - 1}}$, the third will add up to $\frac 4 {8^p} = \frac 1 {2^{4 p - 1}}$, and so on. If $1 < p$, all these quantities are different; in fact, they are getting smaller and smaller. This suggests that possibly, the series doesn’t diverge at all—maybe it converges now instead.

In order to find out whether it converges, we just have to change how we group the terms slightly: we’ll shift them to the left by one term, so that 1 comprises the first group, $\frac 1 {2^p}$ and $\frac 1 {3^p}$ comprise the second group, and so on. Now rather than ending with a power of two, each group begins with a power of two. We will still replace the terms in each group with the power of two in the group, so that each term which changes gets replaced by a larger term. This gives a new series that looks like

$1 + \frac 1 {2^p} + \frac 1 {2^p} + \frac 1 {4^p} + \frac 1 {4^p} + \frac 1 {4^p} + \frac 1 {4^p} + \dotsb.$

The sum of this new series must be larger than the sum of the old one. So if we can prove that the new series has a finite sum, that would mean the old series couldn’t have an infinite sum. In fact, since all the terms in the old series are positive, it would imply something slightly stronger, that the old series must converge to a finite sum (the possibility that the old series has no defined sum at all would be ruled out). I’ll explain how this leap of logic can be made in a later post, since the reasoning required for it is slightly more complex than what’s required for most of the content in this post.

We can, indeed, prove that this new series converges. Let’s add up the terms in each group again. The first group is just 1. The second group has 2 instances of $\frac 1 {2^p}$, which add up to $\frac 2 {2^p} = \frac 1 {2^{p - 1}}$. The third group has 4 instances of $\frac 1 {4^p}$, which add up to $\frac 4 {4^p} = \frac 1 {4^{p - 1}}$. So the resulting series looks like

$1 + \frac 1 {2^{p - 1}} + \frac 1 {4^{p - 1}} + \frac 1 {8^{p - 1}} + \dotsb,$

which can also be written as

$1 + 2^{1 - p} + (2^{1 - p})^2 + (2^{1 - p})^2 + \dotsb$

Perhaps you can see why I’ve written the series in this way: it’s just a geometric series with the common ratio $2^{1 - p}$! Since $1 < p$ so $1 - p < 0$, $2^{1 - p} < 1$, so this series has a finite value equal to $\frac 1 {1 - 2^{1 - p}}$. So now we can conclude that the original series has a finite value no greater than $\frac 1 {1 - 2^{1 - p}}$, which can also be written as $\frac {2^{p - 1}} {2^{p - 1} - 1}$, if we multiply the top and bottom of the fraction by $2^{p - 1}$.

Oh, and I almost forgot to say: what, then, is the finite value of

$1 + \frac 1 {2^p} + \frac 1 {3^p} + \frac 1 {4^p} + \dotsb$

then? This is a considerably harder question to answer. In fact, there isn’t really any way to express the value in a simpler manner than the above series. However, there is a fairly easy way to show that the series can be expressed as an infinite product instead. If we let the value of the series be denoted $\zeta(p)$ (this is standard notation; $\zeta$ is called the zeta function), then it can be seen that

$\frac 1 {2^p} \zeta(p) = \frac 1 {2^p} + \frac 1 {4^p} + \frac 1 {6^p} + \frac 1 {8^p} + \dotsb$

which is just the original series with all the terms with an odd denominator omitted. That means if we subtract this from the original series, we get the series all the terms with an even denominator omitted.

$\left( 1 - \frac 1 {2^p} \right) \zeta(p) = 1 + \frac 1 {3^p} + \frac 1 {5^p} + \frac 1 {7^p} + \dotsb$

Now we can divide this through by $\frac 1 {3^p}$ to get rid of all the terms whose denominators are not multiples of 3. Then subtracting this from $\left( 1 - \frac 1 {2^p} \right) \zeta(p)$ gives us the original series with all the terms whose denominators are multiples of 2 or 3 omitted. We can continue in the same manner, dividing through by $\frac 1 {q^p}$ for every prime number $q$, to get rid of all the terms except the first one.

$\left( 1 - \frac 1 {2^p} \right) \left( 1 - \frac 1 {3^p} \right) \left( 1 - \frac 1 {5^p} \right) \left( 1 - \frac 1 {7^p} \right) \zeta(p) \dotsb = 1$

A simple rearrangement allows us to express $\zeta(p)$ as an infinite product.

$\begin{array}{rll} \zeta(p) &= \left( \frac 1 {1 - 2^{-p}} \right) \left( \frac 1 {1 - 3^{-p}} \right) \left( \frac 1 {1 - 5^{-p}} \right) \left( \frac 1 {1 - 7^{-p}} \right) \dotsb \\ &= \frac {2^p} {2^p - 1} \frac {3^p} {3^p - 1} \frac {5^p} {5^p - 1} \frac {7^p} {7^p - 1} \dotsb \end{array}$

It doesn’t really feel like we have found out what, say, $\zeta(2)$ is if all we can say is that it’s equal to $\frac 4 3 \frac 9 8 \frac 25 24 \dotsb$ as well as $1 + \frac 1 4 + \frac 1 9 + \frac 1 16 + \dotsb$ Still, this is an interesting identity. Also, since we know that $\zeta(1)$ is infinite, this tells us that the infinite product

$2 \frac 3 2 \frac 5 4 \frac 7 6 \dotsb$

is infinite. Now, since the numerators in this infinite product are simply the prime numbers, this yields a proof that there are infinitely many prime numbers. Because if there was a finite number of primes, $2 \cdot 3 \cdot 5 \cdot 7 \dotsb$ would be a finite product, and so would $2 \cdot 4 \cdot 6 \dotsb$ (the product of all numbers one less than a prime). Therefore the ratio $\frac {2 \cdot 3 \cdot 5 \cdot 7 \dotsb} {2 \cdot 4 \cdot 6 \dotsb}$ of these two products would be finite too, but this ratio can also be written as the product above, which we know is equal to the harmonic series and hence infinite.