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I think theirs somthing wrong with this proof as it was not hard to create, if someone could find a mistake I would greatly appreiciate it:

Define a function $[k\equiv b \bmod a]$, to be equal to zero if $k$ isn't congruent to $b \mod a$, and 1 if it is.

From that definition we have:

$$\sum _{k=1}^{\infty }\frac{\ln(k)}{k^s}[k\equiv b.mod. a]=\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k^s}\sum _{j=1}^{\infty} \frac{[jk\equiv \bmod a]}{j^s}$$

(assume the above statement is true ^, its the only lemma I will ask for, and the proof takes 2 long to show here)

But we can brake that sum into parts, with the identity, $$\sum _{j=1}^{\infty}f(j)=\sum_{r=1}^{a}\sum_{j=0}^{\infty}f(aj+r)$$ (were just breaking it up into congruences which still form a basis for all the integers, for example the even and odd integers are the case where a=2)

so we have $$\sum _{k=1}^{\infty }\frac{\ln(k)}{k^s}[k\equiv \bmod a]=\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k^s}\sum _{j=1}^{\infty} \frac{[jk\equiv b.mod. a]}{j^s}$$$$=\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k^s}\sum_{r=1}^{a}\sum _{j=0}^{\infty} \frac{[(aj+r)k\equiv b.mod. a]}{(aj+r)^s}=\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k^s}\sum_{r=1}^{a}\sum _{j=0}^{\infty} \frac{[rk\equiv b.mod. a]}{(aj+r)^s}$$

but also note $\sum_{j=1}^{\infty}\frac{1}{(aj+r)^s}=\frac{\zeta(s)}{a^s}+o(1)$, so we get

$$\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k^s}\sum_{r=1}^{a}\sum _{j=0}^{\infty} \frac{[rk\equiv \bmod a]}{(aj+r)^s}=\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k^s}\sum_{r=1}^{a} [rk\equiv \bmod a](\frac{\zeta(s)}{a^s}+o(1))))=(\frac{\zeta(s)}{a^s}+o(1))\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k^s}\sum_{r=1}^{a} [rk\equiv \bmod a]$$

back tracking a bit we have, $$\sum _{k=1}^{\infty }\frac{\ln(k)}{k^s}[k\equiv \bmod a]=(\frac{\zeta(s)}{a^s}+o(1))\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k^s}\sum_{r=1}^{a} [rk\equiv \bmod a]$$ and thus we have, $$\sum _{j=0}^{\infty }\frac{\ln(aj+b)}{(aj+b)^s}=(\frac{\zeta(s)}{a^s}+o(1))\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k^s}\sum_{r=1}^{a} [rk\equiv \bmod a]$$

(sense all the solutions to $k\equiv b$ mod a, take on the form aj+b for all integers 'j')

and so $$\frac{a^s}{\zeta(s)}\sum _{j=0}^{\infty }\frac{\ln(aj+b)}{(aj+b)^s}=(1+o(\frac{a^s}{\zeta(s)}))\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k^s}\sum_{r=1}^{a} [rk\equiv \bmod a]$$

now taking the limit as s->1, we see

$$\infty=\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k}\sum_{r=1}^{a} [rk\equiv \bmod a]=\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k}[k\equiv b.mod. a]+\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k}[2k\equiv b.mod. a]+\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k}[3k\equiv b.mod. a]+...\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k}[(a-1)k\equiv b.mod. a]$$

and in order for a finite sequence of postive terms to diverge, atleast one of the terms must diverge, so picking out any $0<c<a$, we see there must be some c, greater then zero, and less then a, such that for all b,

$$\infty=\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k}[ck\equiv b.mod. a]$$

and sense $ck\equiv b$ mod a, only has solutions if c and a are coprime, we see that $[ck\equiv b.mod. a]=0$, if c,a arn't coprime, and sense our series diverges, we see that the c we chose must be coprime to a.

So to reiderate weve shown that for some c coprime to a, and less then a, we have $$\infty=\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k}[ck\equiv b.mod. a]$$, now sense we havn't explictly defined the integer b, we can make it a multiple of c at this point, say $b=c*d$, thus we have $$\infty=\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k}[ck\equiv cd.mod. a]$$ but sense c is coprime to a, we may cancle it from both sides of the congruence, giving, $$\infty=\sum _{k=1}^{\infty }\frac{\Lambda(k)}{k}[k\equiv d.mod. a]$$ but sense the vonmangoldt sum is a proxyed sum over the prime powers, and any power greater then 2 is neglible we see,

$$\infty=\sum _p\frac{\ln(p)}{p}[p\equiv d.mod. a]$$

and sense if there were a finite number of primes congruent to d mod a, the series wouldn't diverge, we can conclude dirichlets theorem is true.

(I understand the entire proof is based on the first statement, but I can prove that it is true using only some elementry algebra, and some other ideas I have worked on, and although its too long to post here, if interested you can email me, the proof of the first statrement is about a page long, but it can be condensed)

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If $f(x)=1/x$ then $\sum\limits_{k=1}^nf(k)$ is divergent whereas $nf(n)=1$. – Seirios Dec 22 '12 at 7:59
Is $f$ monotonic increasing? – robjohn Dec 22 '12 at 9:53
The question is only about monotonic function. – Seirios Dec 22 '12 at 9:59
@Ethan: Since there were some good answers to the original question, you should revert this question and ask a new question. – robjohn Dec 22 '12 at 10:03
@Q__: Now my answer to the modified question is orphaned :-) I'll have to delete it and move it if Ethan decides to ask the second question again. – robjohn Dec 22 '12 at 11:20
up vote 7 down vote accepted

You say that we can pick b to be a multiple of c, but your choice of c certainly depends on b. You need to justify the statement that one c works for all b. This is the most likely spot for an error I can find.

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I don't understand why cant I just define b to be a multiple of whatever the constant comes out to be? I never directly manipulated b, to begin with. – Ethan Dec 18 '12 at 4:48
It's like subtraction; the equation $b+c=0$ has a solution b for any c, but it's a different b for every c. If we tried to define b to be 2c, after we had solved the equation, then the equation would no longer be true. I know that,s a bad analogy, but it's the same idea. – Brian Rushton Dec 18 '12 at 5:04

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