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In the third century BC, Euclid proved the existence of infinitely many prime numbers. In the 18th century, Leonhard Euler proved a stronger statement: the sum of the reciprocals of all prime numbers diverges. Here, we present a number of proofs of this result.
## The harmonic series

## First proof

# sum_{p} ln left(frac{1}{1-p^{-1}}right)

sum_{p} - ln(1-p^{-1})
& {} = sum_{p} left(frac{1}{p} + frac{1}{2p^2} + frac{1}{3p^3} + cdots right) = left(sum_{p}frac{1}{p} right) + sum_{p} frac{1}{p^2} left(frac{1}{2} + frac{1}{3p} + frac{1}{4p^2} + cdots right)
& {} < left(sum_{p}frac{1}{p} right) + sum_{p} frac{1}{p^2} left(1 + frac{1}{p} + frac{1}{p^2} + cdots right) = left(sum_{p} frac{1}{p} right) + left(sum_{p} frac{1}{p(p-1)} right)
& {} = left(sum_{p} frac{1}{p} right) + C
end{align}
## Second proof

## Third proof

# sum_{i

1}^nunderbrace{int_i^{i+1}frac{dx}x}_{{}<1/i}
^{2}/6) = 0.4977...; in fact it turns out that
## Fourth proof

## See also

## References

## External links

First, we describe how Euler originally discovered the result. He was considering the harmonic series

- $sum\_\{n=1\}^infty\; frac\{1\}\{n\}\; =$

He had already used the following "product formula" to show the existence of infinitely many primes.

- $sum\_\{n=1\}^infty\; frac\{1\}\{n\}\; =\; prod\_\{p\}\; frac\{1\}\{1-p^\{-1\}\}$

(Here, the product is taken over all primes p; in the following, a sum or product taken over p always represents a sum or product taken over a specified set of primes, unless noted otherwise.)

Such infinite products are today called Euler products. The product above is a reflection of the fundamental theorem of arithmetic. (Multiply out the right side as you would like to do.) Of course, the above "equation" is not necessary because the harmonic series is known (by other means) to diverge. This type of formal manipulation was common at the time, when mathematicians were still experimenting with the new tools of calculus.

Euler noted that if there were only a finite number of primes, then the product on the right would clearly converge, contradicting the divergence of the harmonic series. (In modern language, we now say that the existence of infinitely many primes is reflected by the fact that the Riemann zeta function has a simple pole at s = 1.)

Euler took the above product formula and proceeded to make a sequence of audacious leaps of logic. First, he took the natural logarithm of each side, then he used the Taylor series expansion for ln(1 − x) as well as the sum of a geometric series:

- $$

for a fixed constant C < 1. Since the sum of the reciprocals of the first n positive integers is asymptotic to ln(n), (i.e. their ratio approaches one as n approaches infinity), Euler then concluded

- $frac\{1\}\{2\}\; +\; frac\{1\}\{3\}\; +\; frac\{1\}\{5\}\; +\; frac\{1\}\{7\}\; +\; frac\{1\}\{11\}\; +\; cdots\; =\; ln\; ln\; (+\; infty).$

It is almost certain that Euler meant that the sum of the reciprocals of the primes less than n is asymptotic to ln(ln(n)) as n approaches infinity. It turns out this is indeed the case; Euler had reached a correct result by questionable means.

The following proof by contradiction is due to Paul Erdős.

Let p_{i} denote the i-th prime number. Assume that the sum of the reciprocals of the primes converges, i.e.

- $sum\_\{i=1\}^infty\; \{1over\; p\_i\}\; <\; infty,.$

Then there exists a positive integer k such that

- $sum\_\{i=k+1\}^infty\; \{1over\; p\_i\}\; <\; \{1over\; 2\}.qquad(1)$

For a positive integer x let M_{x} denote the set of those n in {1, 2, . . ., x} which are not divisible by any prime greater than p_{k}. We will now derive an upper and a lower estimate for the number |M_{x}| of elements in M_{x}. For large x, these bounds will turn out to be contradictory.

Upper estimate:
Every n in M_{x} can be written as n = r m^{2} with positive integers m and r, where r is square-free. Since only the k primes p_{1} . . ., p_{k} can show up (with exponent 1) in the prime factorization of r, there are at most 2^{k} different possibilities for r. Furthermore, there are at most √x possible values for m. This gives us the upper estimate

- $|M\_x|\; le\; 2^ksqrt\{x\},.qquad(2)$

Lower estimate:
The remaining x − |M_{x}| numbers in the set difference {1, 2, . . ., x} M_{x} are all divisible by a prime greater than p_{k}. Let N_{i,x} denote the set of those n in {1, 2, . . ., x} which are divisible by i-th prime p_{i}. Then

- $\{1,2,ldots,x\}setminus\; M\_xsubsetbigcup\_\{i=k+1\}^infty\; N\_\{i,x\};.$

Since the number of integers in N_{i,x} is at most x/p_{i} (actually zero for p_{i} > x), we get

- $x-|M\_x|\; le\; sum\_\{i=k+1\}^infty\; |N\_\{i,x\}|<\; sum\_\{i=k+1\}^infty\; \{xover\; p\_i\},.$

Using (1), this implies

- $\{xover\; 2\}\; <\; |M\_x|,.qquad(3)$

Contradiction:
For every integer x ≥ 2^{2k + 2}, the estimates (2) and (3) cannot hold simultaneously.

Here is another proof that actually gives a lower estimate for the partial sums; in particular, it shows that these sums grow at least as fast as ln(ln(n)). The proof is an adaptation of the product expansion idea of Euler. In the following, a sum or product taken over p always represents a sum or product taken over a specified set of primes.

The proof rests upon the following four inequalities:

- Every positive integer i can be uniquely expressed as the product of a square-free integer and a square. This gives the inequality

- $sum\_\{i=1\}^n\{frac\{1\}\{i\}\}\; le\; prod\_\{p\; le\; n\}\{biggl(1+frac\{1\}\{p\}biggr)\}sum\_\{k=1\}^n\{frac\{1\}\{k^2\}\},,$

- The upper estimate for the natural logarithm

- $$

- The lower estimate 1 + x < exp(x) for the exponential function, which holds for all x > 0.
- The upper bound (using a telescoping sum) for the partial sums (convergence is all we really need)

- $$

Combining all these inequalities, we see that

- $$

- $ln\; ln(n+1)\; -\; ln2\; <\; sum\_\{p\; le\; n\}\{frac\{1\}\{p\}\}$

Using

- $$

- $lim\_\{n\; to\; infty\; \}\; biggl($

From Dusart's inequality (see PNT), we get

- $p\_n\; <\; n\; ln\; n\; +\; n\; ln\; ln\; n\; quadmbox\{for\; \}\; n\; ge\; 6.$

Then

- $$

- William Dunham
*Euler The Master of Us All*. MAA.

- Chris K. Caldwell: "There are infinitely many primes, but, how big of an infinity?", http://www.utm.edu/research/primes/infinity.shtml

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Last updated on Tuesday September 30, 2008 at 19:44:44 PDT (GMT -0700)

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Last updated on Tuesday September 30, 2008 at 19:44:44 PDT (GMT -0700)

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