## Thursday, August 31, 2006

### 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/ns. 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

#### 1 comment:

Tony Baker said...

Here is a proof that I did in my high school (not very sure if it will be shorter than the given one)

(1) lemma \Sum a_n is convergent if and only if \Product (1+a_n) is convergent

suppose \Sum 1/p is convergent then \Sum 1/(p-1) is also convergent
Using the lemma \Product (1 + 1/(p-1)) is convergent

1/p-1 = \Sum (n=1->infinity) 1/p^n

So \Product (\Sum (n=0->infinity) 1/p^n) is convergent

however \Product (\Sum (n=0->infinity) 1/p^n) = \Sum (n=1->infinity) 1/n

and we know that \Sum (n=1->infinity) 1/n is infinity --- contradiction

QED