New formulae for the Riemann Zeta Series Back in late April I arrived at the following formula for when $x+b < -1$:
$$\tag{1}\boxed{
 \sum_{a=1}^{\infty}\sum_{k=0}^{\infty}\frac{(k+a)^{x+b}}{a} = \sum_{k=0}^{\infty}\sum_{g=0}^{k}\frac{b^{g}}{(g!)^{2}}\binom{k}{g}\sum_{a=1}^{\infty}\frac{(k+a-g)^{x}}{a}\ln^{g}(k+a-g). 
}$$
I presented my work to a professor and he did not deny its validity.
Now, plug in $0$ for $b$ and notice that the formula is true, though assuming $0^{0}=1$. What is weird is that I then applied the Abel partial summation formula to obtain a new result. I do not see if there is a mistake in my argumentation but the result is either false or extremely counterintuitive. The professor said that the next boxed equation is numerically false.
$$ \sum_{g=0}^{k}\frac{b^{g}}{(g!)^{2}}\binom{k}{g}\sum_{a=1}^{\infty}\frac{(k+a-g)^{x}}{a}\ln^{g}(k+a-g)$$ $$ = \sum_{g=0}^{k}\frac{b^{g}}{(g!)^{2}}\binom{k}{g}\sum_{a=1}^{\infty}\frac{(k+a-g)^{x}}{a}\ln^{-(k-g)}(k+a-g)\ln^{k}(k+a-g), $$
so by Abel's partial summation formula
$$ \sum_{g=0}^{k}\frac{b^{g}}{(g!)^{2}}\binom{k}{g}\sum_{a=1}^{\infty}\frac{(k+a-g)^{x}}{1}\ln^{-(k-g)}(k+a-g)\ln^{k}(k+a-g) $$ $$ = \frac{b^{k}}{k!}\ln^{k}(a)\sum_{g=0}^{k}\frac{1}{g!}\binom{k}{g}(k+a-g)^{x}\ln^{-(k-g)}(k+a-g) $$ $$ -\sum_{n=0}^{k-1}(\frac{b^{n+1}}{(n+1)!}\ln^{n+1}(a+1) - \frac{b^{n}}{n!}\ln^{n}(a))\sum_{g=0}^{n}\frac{1}{g!}\binom{k}{g}(k+a-g)^{x}\ln^{-(k-g)}(k+a-g). $$
I now substitute that result into the equation at the top, set $b=0$ and obtain
$$\tag{2}\boxed{
 (\sum_{a=1}^{\infty}\sum_{k=0}^{\infty}\frac{(k+a)^{x+b}}{a}) - \zeta(-x-b+1) = \sum_{a=1}^{\infty}\sum_{k=1}^{\infty}\frac{(k+a)^{x+b}}{a\ln^{k}(k+a)}. }$$
For $x+b=-2$ the left-hand side of the equation yields $\zeta(3)$. However, the right-hand side of the equation is, if at all, getting there so slowly that a C++ program I wrote never arrives close to that value for quite large upper limits of those sums.
Is it possible for the last equation to be valid? Notice that as $x$ goes to infinity, the equation goes to $0=0$. Moreover, both sides are always positive.
 A: Not an answer, but I've done a bit of work on the last proposed formula.

The last formula you wrote is essentially
$$-\zeta(1-s)+\sum_{m=1}^\infty\sum_{n=0}^\infty\frac{(m+n)^s}{m}=\sum_{m=1}^\infty\sum_{n=1}^\infty \frac{(m+n)^s}{m\log(m+n)^n}$$
Rewrite the LHS slightly as
$$-\zeta(1-s)+\sum_{m=1}^\infty \frac{m^s}{m}+\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(m+n)^s}{m}=\sum_{m=1}^\infty\sum_{n=1}^\infty \frac{(m+n)^s}{m\log(m+n)^n}$$
Recognize the new sum as $\zeta(1-s)$ and combine the remaining bits:
$$\sum_{m=1}^\infty\sum_{n=1}^\infty \frac{(m+n)^s}{m}\left(1-\frac{1}{\log(m+n)^n}\right)=0$$
When $s\in \Bbb{R}$ all of the terms are real. Further, all of the terms are positive, except for $m=n=1$. So now, it remains to show that
$$2^s\left(\frac{1}{\log(2)}-1\right)=\sum_{n=2}^\infty (n+1)^s\left(1-\frac{1}{\log(n+1)^n}\right)+\sum_{m=2}^\infty \frac{(m+1)^s}{m}\left(1-\frac{1}{\log(m+1)}\right)+\sum_{m=2}^\infty\sum_{n=2}^\infty \frac{(m+n)^s}{m}\left(1-\frac{1}{\log(m+n)^n}\right)$$
