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Today, there are a number of proofs of the existence of infinite number of prime numbers,
the first of which was
given by Euclid.
I will give a short proof
of the same fact of my own, and then another proof of not so well known fact about the
prime numbers
(first proven by Euler in 1737) I came up in 1994.
Let's look at the share of non-prime numbers among the natural (positive) numbers. Even numbers
(i.e. numbers divisible by 2) have a share of 1/2: half of natural numbers are even.
In addition, numbers divisible by 3 have a share of 1/3, but taken together with even numbers,
their share would be not 1/2+1/3, but rather: 
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since half of the numbers divisible by 3 are also even, and we don't want to count them twice. Among the numbers
that are divisible by 5, also half are even, some are divisible by 3 and there are some divisible by 2,3 and 5 at the
same time. Therefore, the share of numbers divisible by either 2,3 or 5 will be:
The way to extend these sequence is clear: add 1/Pk, where Pk is the next prime (such as 7,11,13 and so on) and subtract S/Pk, where S is the previously calculated sum:
It is easy to notice that with each step, increment (1 - S(Pk-1))/Pk gets smaller and smaller: 1 - S(Pk-1) decreases, while Pk increases. At the same time, increments of S are non-negative: each time a fraction 1/Pk of remaining (1-S) gets added. Now let's assume that the number of primes is finite, and Pn is the biggest prime. In this case, our sequence is also finite, and Sn should become equal to 1. If it is less than 1 (it cannot be more than one by definition), it means there are infinite counts of natural numbers that are bigger than Pn that are not primes (remember, we assumed that Pn is the last one), but are not non-primes (all non-primes are already included in Sn). By definition, such numbers don't exist, therefore Sn should be equal to 1. It means adding to the current S(Pn-1) 1/Pn-1 and subtracting S/Pn-1 we must get 1. As result, we get this equation: 
It is obvious that if S(n-1) is not equal to 1, this condition cannot be met. If S(n-1) is 1, we go one step backward, and conclude than S(n-2) = 1, using the same logic, and so on. But as we know, not all S(k) are equal to 1, for example S(2) = 1/2. We come to the contradiction, which proves that the original assumption was wrong. Now, using the same representations, I will prove that the harmonic series of primes, (i.e., sum of the reciprocals of the primes) diverges: 
       therefore the original sum won't change and we can say that   
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