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?(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Erdos conjecture on arithmetic progression

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