Sum of the reciprocal of the primes

In today's blog, I will show that the sum of the reciprocal of the primes is divergent, that is, it does not have a limit.

This was first proved by Leonhard Euler. The details in today's blog are based on Euler's original proof.

Lemma 1: ∑ (p=all primes) ∑ (n=2,∞) (p^-n)/n is convergent, that is, it has a finite limit.

Proof:

(1) ∑ (p = all primes) ∑ (n=2,∞) (p^-n)/n is less than ∑ (p = all primes) ∑ (n=2,∞) (p^-n)

This is clear since n is an integer that starts at 2 and goes on until infinity. In each case, we have (p^-n) is greater than (p^-n)/n

(2) Using the infinite geometric series equation (see Lemma 1, here), we know that:

1 + 1/p + 1/p² + ... = 1/(1 - 1/p)

(3) Now (1/p²)*[1 + 1/p + 1/p² + 1/p³ + ... ] = [1/p² + 1/p³ + ..] so that:

∑ (n=2, ∞) (p^-n) = (1/p²)[1/(1 - 1/p)] = 1/[p²(1 - p^-1)]

(4) We can then use step #3 to give us:

∑ (p = all primes) ∑ (n=2, ∞) (p^-n) = ∑ (p = all primes) 1/[p²(1 - p^-1)] =

= ∑ (p = all primes) 1/[(p)(p)(1 - p^-1)] =

= ∑ (p = all primes) 1/[(p)(p - 1)]

(5) ∑ (p = all primes) 1/[(p)(p-1)] is less than ∑ (n=2, ∞) 1/[(n)(n-1)] since n includes nonprimes in the sum.

(6) ∑ (n=2, ∞) 1/[(n)(n-1)] = ∑ (n=2,∞) 1/(n² - n) is less than ∑ (n=2, ∞) 1/(n² - 2n + 1)

This is true since n² - 2n + 1 is always less than n² - n when n ≥ 2.

(7) Since n² - 2n + 1 = (n-1)², we have:

∑ (n=2,∞) 1/(n-1)² = 1 + 1/4 + 1/9 + ... = π²/6 (See Theorem, here)

(8) So, we have now proved that:

∑ (p = all primes) ∑ (n=2,∞) (p^-n)/n is less than π²/6 which is finite.

QED

Theorem: Sum of the Reciprocal of Primes is Divergent

∑ 1/p = 1/2 + 1/3 + 1/5 + 1/7 + ... is divergent, that is, does not have a limit.

Proof:

(1) We know that the ζ(1) = harmonic series, that is, ∑ 1/n diverges, that is, the limit of ∑ 1/n approaches infinity. [See Lemma 3, here]

NOTE: ζ(s) is the zeta function which is equal to ∑ 1/n^s. So ζ(1) = ∑ 1/n¹ = ∑ 1/n.

(2) From step #1, the limit of log(ζ(1)) approaches infinity. This follows directly from step #1 and the fact that as x gets bigger, log x likewise gets bigger. As x heads to infinity, log x also heads to infinity (see here for background on logarithms)

(3) Using Euler's Product Formula (see Theorem 4, here), we know that:

log(ζ(1)) = log(∑ 1/n) = log(∏ (1 - p^-1)^-1)

NOTE: ∏ (1 - p^-1)^-1) is a product of all primes p.

(4) By the property of logarithms (see here), we have:

log(∏ (1 - p^-1)^-1) = ∑ log([1 - p^-1]^-1) = ∑ -log(1 - p^-1)

(5) Now using some basic properties of the integral (see Theorem, here), we have:

log (1 + x) = ∑ (n=1,∞) [(-1)ⁿ⁺¹xⁿ]/n

Let x = -p^-1

Then we have:

log (1 + [-p^-1]) = ∑ [(-1)ⁿ⁺¹(-p^-1)ⁿ]/n = ∑ [(-1)ⁿ⁺¹(-1)ⁿ(p^-1)ⁿ]/n =

= ∑ [(-1)²ⁿ⁺¹(p^-n)]/n

Since 2n+1 is always odd, this gives us:

log(1 + [-p^-1]) = ∑ -(p^-n)/n

Since p^-n/n is always positive, we have ∑ -(p^-n)/n = (-1)∑ (p^-n)/n

Further,

-log(1 + [-p^-1]) = (-1)log(1 + [-p^-1]) = (-1)(-1)∑ (p^-n)/n = ∑ (p^-n)/n

(6) Combining step #5 with step #4 gives us:

∑ -log(1 - p^-1) = ∑ (p=all primes) ∑ (n=1,∞) (p^-n)/n

(7) ∑ (p=all primes) ∑ (n=1,∞) (p^-n)/n=∑(p=all primes) 1/p + ∑ ∑ (n=2,∞) (p^-n)/n

(8) Using Lemma 1 above, we know that ∑ ∑ (n=2,∞) (p^-n)/n is convergent, that is, it has a finite limit.

(9) But from step #2, we know that ∑ 1/p + ∑ ∑ (n=2, ∞) (p^-n)/n has a limit of infinity.

(10) Therefore, it follows, that ∑ 1/p must be divergent, that is, it must have a limit of infinity.

QED

References

• Jay R. Goldman, The Queen of Mathematics, A K Peters, 1998
• Proof that the Sum of the Reciprocals of Primes Diverge, Wikipedia
• Natural Logarithm, Wikipedia