Divergence of
P
p 1/p
Adrian Dudek
adrian.dudek[at]anu.edu.au
When I was in high school, my maths teacher cheekily told me that it’s possible to add up infinitely
many numbers and get a finite number. She then illustrated this by writing
1+
1 1 1
1
+ + +
+ · · · = 2.
2 4 8 16
My jaw dropped. I asked her how she knew this amazing thing. Here was her response.
Teacher’s Response
We can show that the sum of all those things on the left equals 2 as follows. First, let’s say that
it equals S, and then we shall use some clever maths to show that S is equal to 2. We have:
S =1+
1
1 1 1
+ + +
+ ...
2 4 8 16
Look what happens when we multiply through by 1/2:
1
1 1 1
1
S= + + +
+ ...
2
2 4 8 16
We get the original sum, except we are missing the 1. If we add 1 to both sides we have
1
1 1 1
1
S+1=1+ + + +
+ ...
2
2 4 8 16
But the right hand side of the above equation is the thing we have called S, and so we have
that
1
S+1=S
2
You can then solve this equation to get S = 2.
I walked out of the class with my mouth still open, shocked at what I had just seen.
Later that afternoon, I got out a piece of paper and started to play with this cool technique my
teacher had taught me.
The next day I stuck around after class to tell my teacher the good news. I strode up to the
blackboard and wrote
1
1 + 2 + 4 + 8 + 16 + · · · = −1
She just stared at me blankly. “That’s not quite right, I’m afraid” she said.
“Actually, it is” I retorted. “Check this out!”
What I Wrote On The Blackboard
S = 1 + 2 + 4 + 8 + 16 + . . .
If I multiply both sides by 2 I get
2S = 2 + 4 + 8 + 16 + 32 + . . .
Now, if I add a 1 to both sides I get
2S + 1 = 1 + 2 + 4 + 8 + 16 + 32 + . . .
Wow, I get my original sum S. This means that
2S + 1 = S
I then solve for S and get S = −1.
I grinned at her triumphantly but she didn’t grin back. She simply told me that my working out was
wrong. Her reasoning was that you can’t add up infinitely many things that get bigger. If you want
to add up infinitely many things, they must become really really small quickly so that they become
negligible. She said as long as the numbers get closer to zero, the sum will converge to a finite value.
Alas, in my first year at university, I discovered that she had tricked me again. In a lecture on
infinite series, my professor wrote down the Harmonic Series
1+
1 1 1
+ + + ...
2 3 4
and assured us that it does not converge, that is, it buggers off to infinity. The professor convinced
us of this as follows.
The Harmonic Series Diverges
Let’s write out the harmonic series like so:
1+
1 1 1 1 1 1 1
+ + + + + + + ...
2 3 4 5 6 7 8
Now we insert some brackets, just to group things together.
2
1+
1 1 1 1 1 1 1
+
+
+
+ + +
+ ...
2
3 4
5 6 7 8
Our first set of brackets groups two things, the next set groups four things, the one after will
group eight, and so on in powers of two. Let’s look at the first bracket:
1
3
+
1
4
The smallest of these two things is 1/4, and so we know that
1
+
3
1 1 1 1
>
+
=
4
4 4
2
In the next pair of brackets
1
5
+
1 1 1
+ +
6 7 8
we see that the smallest term is 1/8, and so again we get
1
5
+
1 1 1 1 1 1 1 1
+ +
>
+ + +
=
6 7 8
8 8 8 8
2
Continuing in this fashion, we get that
1+
1 1 1 1 1 1 1
+
+
+
+ + +
+ ...
2
3 4
5 6 7 8
is greater than
1+
1 1 1
+
+
+ ...
2
2
2
which clearly goes off to infinity. As our original sum is bigger than this, our original sum must
go off to infinity as well.
Upon finishing this proof, the professor delivered an even cheekier smile than my high school teacher,
as if he knew that we were all spoon-fed some garbage about terms in the sum getting smaller and
smaller.
I’m a little bit wiser now, and I know that if we want a sum of the form
1
1
1
+
+
+ ...
a1 a2 a3
to converge, the terms need to go to zero at a good pace, or put another way, the denominators
need to go to infinity at a good pace.
For example, the harmonic series
3
1+
1 1 1
+ + + ...
2 3 4
has denominators which are the counting numbers
1, 2, 3, 4, . . .
and sure, these head off to infinity but not very quickly. Another example I’ve seen in my time is
the following:
1+
1 1
1
π2
+ +
+ ··· =
4 9 16
6
Here, the denominators are square numbers, a sequence which grows sufficiently quickly to guarantee
convergence.
Here’s a super cool fact which tells us something about growth in the sequence of primes.
X1
p
p
=
1 1 1 1
1
+ + + +
+ ··· = ∞
2 3 5 7 11
That is, the sequence of prime numbers does not grow quickly enough to guarantee convergence.
In some sense, primes are denser than squares, and this may have prompted Legendre’s conjecture,
the statement that there is a prime between any two square numbers. This conjecture is currently
unproven, in fact, we don’t even know if there is a prime between any two cubes!
We are going to prove that the above series indeed diverges. First, we will need a little lemma.
Lemma 1 Let s be a real number which is greater than 1. Then
1+
1
1
1
1
1
1
1
+ s + s + s + ··· =
.
.
...
s
−s
−s
2
3
4
5
1−2
1−3
1 − 5−s
where the left side is a sum over all counting numbers, and the right side is a product over all
of the prime numbers. Using formal notation, this is
X 1
Y
1
=
s
n
1
−
p−s
n
p
Here’s a somewhat informal proof of the above identity.
4
Informal Proof of Identity
First, notice that each term on the right hand side of the identity is of the form
1
1 − p−s
.
It is known that for |x| < 1 we have the expansion
1
= 1 + x + x2 + x3 + . . . .
1−x
As |p−s | < 1, we have
1
1 − p−s
= 1 + p−s + (p−s )2 + (p−s )3 + . . .
= 1 + p−s + p−2s + p−3s + . . .
It is now clear that
Y
p
1
1 − p−s
can be written as
Y
(1 + p−s + p−2s + p−3s + . . . ).
p
Now let’s think about this product. This might be easier if we write it out as
(1 + 2−s + 2−2s + 2−3s + . . . )(1 + 3−s + 3−2s + 3−3s + . . . )(1 + 5−s + 5−2s + 5−3s + . . . )
If we then expand these brackets completely, we will end up with a sum. Each term in this sum
(except for 1, which we get by multiplying the 1 in each bracket together) is going to be a product
of primes or prime powers, all to the power of −s. Since every integer greater than 1 is a unique
product of primes and prime powers, we are going to recover every positive integer here exactly
once and to the power of −s:
1 + 2−s + 3−s + 4−s + . . .
This is exactly equal to
P
n 1/n
s,
which is the left hand side of our identity!
The above proof is informal in the sense of manipulating infinite sums and products. One can make
5
it formal by starting with a product of the first n primes and applying the Taylor series from there.
Then, you show that the difference between this and the infinite sum tends to zero as n → ∞.
P
Now, let’s prove that p 1/p = ∞.
The Reciprocal Prime Series Diverges Let s > 1. We start with the identity:
X 1
Y
1
=
s
n
1
−
p−s
n
p
Taking the natural log of both sides we get
log
Y
X 1 1 =
log
ns
1 − p−s
n
p
1 X
log
=
1 − p−s
p
X
= −
log(1 − p−s )
p
We now wish to use the following Taylor series, which holds for |x| < 1.
− log(1 − x) =
∞
X
xn
n=1
n
We throw this into the ring to get
log
X 1 X
=
− log(1 − p−s )
s
n
n
p
X 1
1
1
=
+
+
+
·
·
·
ps 2p2s 3p3s
p
X 1
X 1 1
1
1
=
+
+
+
+
·
·
·
ps
p2s 2 3ps 4p2s
p
p
X 1
X 1 1
1
<
+
1
+
+
+
·
·
·
ps
p2s
ps p2s
p
p
We then use our result on summing an infinite geometric series to get:
log
X 1 X 1
X
1
<
+
s
s
s
s
n
p
p (p − 1)
n
p
p
Let’s consider the limit as s → 1. The left side will tend to infinity as we know that the harmonic
series will go off to infinity.
6
As the right side is greater than the left side, the right side must also go off to infinity. Consider
the sum
1
X
p
ps (ps
− 1)
.
We can see that
X
p
1
ps (ps − 1)
<
X 1
p2s
p
∞
X
1
<
n2s
n=1
As s → 1, the above sum is bounded by π 2 /6, as mentioned earlier. Going back to our main
equation:
log
X 1 X 1
X
1
<
+
s
s
s
s
n
p
p (p − 1)
p
p
n
We see that
X 1
ps
p
is forced to go to infinity as s → 1, for the left hand side heads to infinity and the other sum
over the prime numbers is bounded.
Therefore,
X1
p
p
7
=∞