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