Infinitude of Primes
Via Harmonic Series
It's well known that the harmonic series diverges. We may thin it somewhat by removing all terms with denominators divisible by 2. What is left over is the series of the reciprocals of the odd whole numbers:
(1) | Σ_{i = 1} 1/(2i - 1). |
This series is divergent:
Σ_{i = 1} 1/(2i - 1) > Σ_{i = 1} 1/2i = ½Σ_{i = 1} 1/i,
which should be understood as a manipulation of partial sums of the two series. The partial sums of the reciprocals of the odd numbers are greater than the corresponding sums of the harmonic series divided by 2. Since the latter diverge, so do the former.
Next we exclude from (1) the terms with the denominators that are multiples of 3. The remaining terms are the reciprocals of the integers in one of the forms: either
We may continue this process indefinitely removing eventually the reciprocals of all the composite numbers and leaving only those of the primes. The sieve of Eratosthenes of sorts.
I'll use the classic argument from Hardy and Wright to show that the remaining series, i.e. the series of the reciprocals of the primes is still divergent. From here it will follow that the number of primes cannot be finite. (This prooof is also discussed at length in Ingenuity in Mathematics, and a different proof can be found in, say Proofs from THE BOOK.)
Assume all the primes are enumerated according to their magnitude: 2, 3, 5, ..., p_{k} , ..., and consider the first k. For a given x, define N(x) as the number of terms in the set
{n ≤ x: n is not divisible by any prime beyond p_{k}}.
In words: N(x) is the number of n's not exceeding x with only prime divisors p satisfying p ≤ p_{k}. Any n from that set can be represented in the form
n = (n_{1})^{2} m,
with a square-free m:
m = 2^{a1}3^{a2}5^{a3}· ... ·p_{k}^{ak},
where every a is either 0 or 1. There are just 2^{k} possible combinations of the k exponents and so not more than 2^{k} possible values for m.
On the other hand,
n_{1} ≤ √n ≤ √x
and so there are not more than √x different values for n_{1} implying that
N(x) ≤ 2^{k}√x.
If the number of primes is finite and their total is k then
x ≤ 2^{k}√x
or
x ≤ 2^{2k}
which is false for x > 2^{2k}.
We now use the same approach to prove the following
Theorem
Let 2, 3, 5, ..., p_{k} be an enumeration of the primes. Then the series
Σ 1/p_{k} = 1/2 + 1/3 + 1/5 + ... + 1/p_{k} + ...
is divergent.
Proof
If the series convergent then its partial sums tend to a limit so that the remainders
r_{n} = Σ_{k > n} 1/p_{k} = 1/p_{n +1} + 1/p_{n +2} + ...
become arbitrary small. In particular, there is n such that r_{n} < 1/2:
1/p_{n +1} + 1/p_{n +2} + ... < 1/2.
The number of n ≤ x which are divisible by p is at most x/p. Hence,
x/p_{n +1} + x/p_{n +2} + ... < x/2.
From the foregoing discussion we conclude that
x/2 ≤ N(x) ≤ 2^{n}√x,
or, again, a contradictory
x ≤ 2^{2(n + 1)}.
We see that the assumption that the series of the reciprocals of the primes is convergent leads to a contradiction. Therefore, the series diverges.
Reference
- M. Aigner, G. Ziegler, Proofs from THE BOOK, Springer, 2000
- G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth Edition, Oxford Science Publications, 1996.
- R. Honsberger, Ingenuity in Mathematics, MAA, New Math Library, 1970
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