# 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.

• These sums are numerically tricky alright, but I'm consistently getting $\sim 0.602$ for the right hand side of the last proposed equation, which is about half of what you are looking for. Check for any missing factors of $2$. May 25 at 13:17
• @K.defaoite ${\rm RHS} > 0.6025 > \frac{1}{2}\zeta (3) = 0.6010284515 \ldots$
– Gary
May 25 at 13:56
• When you apply Abel summation you divide $(k+a-g)^x$ by $1$ and not $a$, why? Towards the end you say "set $b=0$", but then $b$ appears in your final result.
– Gary
May 25 at 14:02
• I do not need the a factor for that separate computation, notice that I later plug the obtained result to the original formula which has that factor. b is there but I still assume it is 0. May 25 at 14:40
• Please number your equations with \tag{1}, \tag{2} and so on. I can guess what "the next boxed equation" refers to but it is all the way at the end. Referring to equations by numbers avoids ambiguity. May 27 at 22:26

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 $$2^s\left(\frac{1}{\log(2)}-1\right)=\sum_{n=2}^\infty (n+1)^s\left(1-\frac{1}{\log(n+1)^n}\right)+\sum_{m=2}^\infty \frac{(m+1)^s}{m}\left(1-\frac{1}{\log(m+1)}\right)+\sum_{m=2}^\infty\sum_{n=2}^\infty \frac{(m+n)^s}{m}\left(1-\frac{1}{\log(m+n)^n}\right)$$
• I went up to $500$ with both $a$ and $k$ and got about $0.6025$.