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771 ：memo[]：2018/04/13(金) 09:11:02.74 ID:p9cqkdQk - Let's prove that there exists infinitely many primes.
Suppose there exists only n primes.
Π[j=1,n](Σ[i=0,∞]1/(p_j)^i) = (1/1+1/2+1/4+…)(1/1+1/3+1/9+…)(1/1+1/5+1/25+…)…(1/1+1/(p_n)+1/(p_n)^2+…)
where p_j denotes the jth prime.
By the unique factorization theorem, when we expand this, we get every reciprocal of a natural number once. So,
Π[j=1,n](Σ[i=0,∞]1/(p_j)^i) = Σ[k=1,∞]1/k.
Also, {1/(p_j)^i} is a geometric sequence with first term 1 and common ratio 0<1/(p_j)<1. So,
Π[j=1,n](Σ[i=0,∞]1/(p_j)^i) = Π[j=1,n]1/(1-1/(p_j)) = Π[j=1,n](p_j)/((p_j)-1).
We obtain
Σ[k=1,∞]1/k = Π[j=1,n](p_j)/((p_j)-1).
The L.H.S. is the harmonic series known to diverge, while the R.H.S. is a finite number.
Contradiction.
Therefore, there exists infinitely many primes.
Let's prove that the sum of the reciprocals of the primes diverges.
Π[j=1,∞](p_j)/((p_j)-1) = ∞.
For any x, 1+x ≦ e^x, so,
Π[j=1,a](p_j)/((p_j)-1) = Π[j=1,a]1+1/((p_j)-1) ≦ Π[j=1,a]e^(1/((p_j)-1)) = e^(Σ[j=1,a]1/((p_j)-1)).
Thus
e^(Σ[j=1,∞]1/((p_j)-1)) = ∞
and
Σ[j=1,∞]1/((p_j)-1) = ∞
follows.
For j≧2, (p_j)-1 ≧ p_(j-1), and 1/((p_j)-1) ≦ 1/(p_(j-1)), so,
Σ[j=1,b]1/((p_j)-1) = 1/(2-1) + Σ[j=2,b]1/((p_j)-1) ≦ 1 + Σ[j=2,b]1/(p_(j-1)) = 1 + Σ[j=1,b-1]1/(p_j).
Thus
Σ[j=1,∞]1/(p_j) = ∞
WWWWW.
| - 分からない問題はここに書いてね442
365 ：１３２人目の素数さん[]：2018/04/13(金) 16:47:42.57 ID:p9cqkdQk - a=b=c=0を満たす2桁の自然数はない
0<b=2a<6より(a,b,c)=(1,2,4),(2,4,8)だが
4k+1=6l+2は両辺の偶奇が異なり(1,2,4)は不適
10≦4k+2=6l+4=13m+8≦99
mは偶数で2nとおくと26n+8
34,60,86のうち34のみ適
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