Erdos conjecture on arithmetic progression

[jinchuriki300]

I read this through wikipedia and some other sources and find it to be unsolved. Erdos offer a prize of $5000 to prove it. A mathematician at UW has looked at it and verify them to be correct. However, i still have some doubt about it because the proof i give is pretty simple. Can anyone take a look at it?
STATEMENT AND DEFINITION from wikipedia

Formally, if

\sum_{n\in A} \frac{1}{n} = \infty

(i.e. A is a large set) then A contains arithmetic progressions of any given length.

If true, the theorem would generalize Szemerédi's theorem.

Erdős offered a prize of US$3000 for a proof of this conjecture at the time. The problem is currently worth US$5000.

The Green–Tao theorem on arithmetic progressions in the primes is a special case of this conjecture

Proof

For positive arithmetic progression which d ≥ 1

let's say f(x) = 1/n + 1/(n+d) + 1/(n+2d) + ...+1/(n+xd)

then n.f(x) = n/n + n/(n+d) + n/(n+2d) +...+n/(n+xd) = 1 + 1/(1+d/n) + 1/(1+2d/n) +...1/(1+xd/n)

x starts at 1 and goes to infinity, then there are some x that could be divisible by n,

then n.f(x) = 1 + 1/(1+d/n) +...1/(1+d) + 1/(1+xd/n) +...1/(1+2d)+...

n.f(x) = a(x) + b(x) + c(x) +...

a(x) = 1 + 1/(1+d) + 1/(1+2d) +...

hence f(x) = a(x)/n + b(x)/n +...

now, a new function series m(x) = a(x) - 1 = 1/(1+d) + 1/(1+2d) +...

defined a new function n(x) = 1/(d+d) + 1/(d+2d) + 1/(d+3d) +... = 1/2d + 1/3d + 1/4d +...= (1/2 + 1/3 + 1/4 + 1/5+...)/d

Thus, (1/2 + 1/3 + 1/4 + 1/5+...) diverges so does (1/2 + 1/3 + 1/4 + 1/5+...)/d, then n(x) diverges, m(x) > n(x), hence m(x) diverges, m(x) + 1 diverges, therefore a(x) diverges, that mean a(x)/n diverges for all n, then f(x) diverges.

For negative arithmetic progression which d ≥ 1 ; d > n

let's say f(x) = 1/n + 1/(n-d) + 1/(n-2d) + ...+1/(n-xd)

Using the same method above: f(x) = (1+ 1/(1-d) + 1/(1-2d) + 1/(1-3d) + ...)/n + b(x)/n +...

a(x)=1 + 1/(1-d) + 1/(1-2d) + 1/(1-3d) + ... = negative result =∞ when d = 1, therefore we have to prove d > 1

Now, m(x) = a(x) - 1 = 1/(1-d) + 1/(1-2d) + 1/(1-3d) + ...

n(x) ≥ m(x)

because n(x) = 1/(-d-d) + 1/(-d-2d) + ...= 1/(-2d) + 1/(-3d) + ... = - (1/2+1/3+1/4+...)/d = ∞

Hence f(x) diverges

now we have proved d ≥ 1

But considered f(x) = 1/(n+y) + 1/(n+2y) and g(x) = 1/(n+t) + 1/(n+2t) +...

if y > t and f(x) diverges, then g(x) diverges. So 1 ≥ d is true if d ≥ 1 is true

 

[Norwegian]

Hi, in your proof, I read various statements to the effect that the harmonic (1+1/2+1/3+1/4+...) and similar series are divergent. I do not see anything related to the properties of a given large set A.

It is interesting to note that the Erdos-conjecture you mention is not even known for arithmetic sequences of length 3. I really think someone should finally settle that.

 

[jinchuriki300]

what do you mean by length of 3?

 

[ramsey2879]

I read this through wikipedia and some other sources and find it to be unsolved. Erdos offer a prize of $5000 to prove it. A mathematician at UW has looked at it and verify them to be correct. However, i still have some doubt about it because the proof i give is pretty simple. Can anyone take a look at it?
STATEMENT AND DEFINITION from wikipedia

Formally, if

\sum_{n\in A} \frac{1}{n} = \infty

(i.e. A is a large set) then A contains arithmetic progressions of any given length.

If true, the theorem would generalize Szemerédi's theorem.

Erdős offered a prize of US$3000 for a proof of this conjecture at the time. The problem is currently worth US$5000.

The Green–Tao theorem on arithmetic progressions in the primes is a special case of this conjecture

Proof

For positive arithmetic progression which d ≥ 1

let's say f(x) = 1/n + 1/(n+d) + 1/(n+2d) + ...+1/(n+xd)

then n.f(x) = n/n + n/(n+d) + n/(n+2d) +...+n/(n+xd) = 1 + 1/(1+d/n) + 1/(1+2d/n) +...1/(1+xd/n)

x starts at 1 and goes to infinity, then there are some x that could be divisible by n,

then n.f(x) = 1 + 1/(1+d/n) +...1/(1+d) + 1/(1+xd/n) +...1/(1+2d)+...

n.f(x) = a(x) + b(x) + c(x) +...

a(x) = 1 + 1/(1+d) + 1/(1+2d) +...

hence f(x) = a(x)/n + b(x)/n +...

now, a new function series m(x) = a(x) - 1 = 1/(1+d) + 1/(1+2d) +...

defined a new function n(x) = 1/(d+d) + 1/(d+2d) + 1/(d+3d) +... = 1/2d + 1/3d + 1/4d +...= (1/2 + 1/3 + 1/4 + 1/5+...)/d

Thus, (1/2 + 1/3 + 1/4 + 1/5+...) diverges so does (1/2 + 1/3 + 1/4 + 1/5+...)/d, then n(x) diverges, m(x) > n(x), hence m(x) diverges, m(x) + 1 diverges, therefore a(x) diverges, that mean a(x)/n diverges for all n, then f(x) diverges.

For negative arithmetic progression which d ≥ 1 ; d > n

let's say f(x) = 1/n + 1/(n-d) + 1/(n-2d) + ...+1/(n-xd)

Using the same method above: f(x) = (1+ 1/(1-d) + 1/(1-2d) + 1/(1-3d) + ...)/n + b(x)/n +...

a(x)=1 + 1/(1-d) + 1/(1-2d) + 1/(1-3d) + ... = negative result =∞ when d = 1, therefore we have to prove d > 1

Now, m(x) = a(x) - 1 = 1/(1-d) + 1/(1-2d) + 1/(1-3d) + ...

n(x) ≥ m(x)

because n(x) = 1/(-d-d) + 1/(-d-2d) + ...= 1/(-2d) + 1/(-3d) + ... = - (1/2+1/3+1/4+...)/d = ∞

Hence f(x) diverges

now we have proved d ≥ 1

But considered f(x) = 1/(n+y) + 1/(n+2y) and g(x) = 1/(n+t) + 1/(n+2t) +...

if y > t and f(x) diverges, then g(x) diverges. So 1 ≥ d is true if d ≥ 1 is true

I don't see where you taken an arbitary infinite random set of numbers and shown that if the sum of the recipricals diverges i.e. equals infinity then the random set of numbers contains arithmetic progressions of any given length. It looks like you chose specific arithmetic sequences, not infinite random sets of numbers.
See
http://en.wikipedia.org/wiki/Erdős_conjecture_on_arithmetic_progressions
for a statement of the conjecture

 

[blue_raver22]

I appreciate your efforts in attempting to solve the Erdos conjecture on arithmetic progression. It is a challenging problem that has yet to be proven, despite many attempts by mathematicians. While your proof may seem simple, it is important to remember that a proof must be rigorous and withstand scrutiny from other mathematicians. I encourage you to share your proof with other experts in the field and see if they can find any flaws or offer suggestions for improvement. Only with a strong and indisputable proof can the Erdos conjecture be officially solved and the prize claimed. Keep exploring and pushing the boundaries of mathematical knowledge – who knows, you may be the one to finally crack the Erdos conjecture!