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.

  • $\begingroup$ 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 $2$. $\endgroup$
    – K.defaoite
    May 25 at 13:17
  • $\begingroup$ @K.defaoite ${\rm RHS} > 0.6025 > \frac{1}{2}\zeta (3) = 0.6010284515 \ldots$ $\endgroup$
    – Gary
    May 25 at 13:56
  • $\begingroup$ 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. $\endgroup$
    – Gary
    May 25 at 14:02
  • $\begingroup$ 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. $\endgroup$ May 25 at 14:40
  • $\begingroup$ 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. $\endgroup$
    – Somos
    May 27 at 22:26

1 Answer 1


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 $$2^s\left(\frac{1}{\log(2)}-1\right)=\sum_{n=2}^\infty (n+1)^s\left(1-\frac{1}{\log(n+1)^n}\right)+\sum_{m=2}^\infty \frac{(m+1)^s}{m}\left(1-\frac{1}{\log(m+1)}\right)+\sum_{m=2}^\infty\sum_{n=2}^\infty \frac{(m+n)^s}{m}\left(1-\frac{1}{\log(m+n)^n}\right)$$

  • $\begingroup$ Please use the notation that has powers of logarithms before their arguments so as to make it less confusing. Furthermore, I was about to say that I'm also getting about 0.602 out of my program but I do not see any terms missing in my derivation. But notice that it makes quite a difference if you compute only the first few terms of the k sum for every iteration of the a sum. With certain chocies you would think you could never get beyond 0.6. $\endgroup$ May 25 at 13:47
  • $\begingroup$ I went up to $500$ with both $a$ and $k$ and got about $0.6025$. $\endgroup$
    – Gary
    May 25 at 14:00
  • $\begingroup$ Try 50000 for a and only 5 for k and you will get less than 0.6. $\endgroup$ May 25 at 14:41
  • $\begingroup$ You must have made a mistake somewhere. There LHS is negative. $\endgroup$ May 26 at 11:56
  • 1
    $\begingroup$ @ArturWiadrowski Yes sorry, forgot to flip the sign when moving it over thanks $\endgroup$
    – K.defaoite
    May 26 at 14:30

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