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Out of curiosity, I was trying to generate an identity involving $\zeta(s)$ and $\zeta(s+1)$, for integers $s>1$, and after a lot of scribbling ended up with the following:

$(\zeta(s)-1)*(\zeta(s+1)-1)=\sum_{c=4}^\infty((\sum \frac 1f)*\frac 1{c^s})$, where the outer summation runs for all composites $c\geq4$ and the inner summation is for all the factors $f$ of $c$ (excluding $1$ and $c$ themselves).

Simply put, the entire RHS would look like: $(\frac 12)*\frac 1{4^s}+(\frac 12+\frac 13)*\frac 1{6^s}+(\frac 12+\frac 14)*\frac 1{8^s}+(\frac 13)*\frac 1{9^s}+(\frac 12 + \frac 15)*\frac 1{10^s}+(\frac 12 + \frac 13 + \frac 14 + \frac 16)*\frac 1{12^s}+...$

Now, before I ask my question, I must tell you that I don't have formal training or an advanced degree in mathematics. My question:

Can the above identity be derived starting only with either LHS or RHS? I ask this because my derivation involves construction of a variation of $\zeta$, doing it for all the numbers upto a number $N$, adding those constructions, and then taking $\lim_{N\to\infty}$ (I can share the entire derivation if asked). I also want to know if there is something obviously intuitive about this identity that I am missing (that led me to deriving it the long way).

Also if somebody could point me to some literature that deals with similar identities, I would be highly grateful. Thanks!

PS: I tested the identity with $s=2$ (therefore using $\zeta(2)=\frac {\pi^2}6$ and the value of $\zeta(3)$ from https://oeis.org/A002117 upto around a hundred decimal places), and with the composites $c$ upto 10,000 for the outer summation. The LHS and RHS values do match to 4 decimal places. I am sure adding more composites to the summation will give more accurate results.

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    $\begingroup$ $\zeta(s+1)\zeta(s)=\sum_{n\ge 1}n^{-s}\sum_{d|n} \frac1d$ then substract $\zeta(s)+\zeta(s+1)=\sum_{n\ge 1}n^{-s} (1+\frac1n)$ to get $(\zeta(s+1)-1)(\zeta(s)-1)=\sum_{n\ge 1}n^{-s}\sum_{d|n,1<d<n} \frac1d$ $\endgroup$
    – reuns
    Jun 29, 2020 at 4:07
  • $\begingroup$ @reuns That was quick! I am having trouble wrapping my head around your first identity itself and need help in deriving it. Can you please elaborate? $\endgroup$ Jun 29, 2020 at 14:03
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    $\begingroup$ The first identity comes from $\zeta(s)\zeta(s+1) = \sum \frac{1}{n^s m^{s+1}} = \sum \frac{1}{(nm)^s} \frac{1}{m}$. Then sum over $N = nm$ instead and get $\sum_N \frac{1}{N^s} \sum_{d \mid N} \frac{1}{d}$. $\endgroup$
    – davidlowryduda
    Jun 30, 2020 at 16:44
  • $\begingroup$ Thank you @davidlowryduda for the clarification. I want to mark reuns's comment as the answer, but can't figure out how. $\endgroup$ Jun 30, 2020 at 17:25
  • $\begingroup$ Ah, well both reuns and I are doing a somewhat poor habit of giving most of an answer in comments. But comments should be considered temporary and ephemeral, and are possibly deleted from time to time without notice. I'll put these together into an answer now. $\endgroup$
    – davidlowryduda
    Jun 30, 2020 at 17:33

1 Answer 1

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Through direct expansion, we compute that $$ \zeta(s+1)\zeta(s) = \sum_{n \geq 1} \sum_{m \geq 1} \frac{1}{n^sm^{s+1}} = \sum_{n \geq 1} \sum_{m \geq 1} \frac{1}{(nm)^s} \frac{1}{m} = \sum_{N \geq 1} \frac{1}{N^s} \sum_{d \mid N} \frac{1}{d}.\tag{1}$$ We also have the trivial identity $$ \zeta(s) + \zeta(s+1) = \sum_{n \geq 1} \Big(\frac{1}{n^s} + \frac{1}{n^{s+1}}\Big) = \sum_{n \geq 1} \frac{1}{n^s} \Big( 1 + \frac{1}{n}\Big).\tag{2}$$ On the one hand, we recognize the product you're looking for as $$ \zeta(s+1)\zeta(s) - \zeta(s) - \zeta(s+1) + 1 = (\zeta(s+1) - 1)(\zeta(s) - 1).$$ We can compute this in terms of $(1)$ and $(2)$ by subtracting $(2)$ from $(1)$. This gives $$ \zeta(s+1)\zeta(s) - \zeta(s) - \zeta(s+1) + 1 = \sum_{n \geq 1} \frac{1}{n^s} \sum_{\substack{d \mid n \\ 1 < d < n}} \frac{1}{d}, $$ as $(1 + 1/n)$ are exactly the terms $1/d$ for $d = 1$ and $d = n$. Note the annoying indexing in exactly the $n = 1$ case, as the coefficient in $\zeta(s+1)\zeta(s)$ is $1/1^s = 1$, while the coefficient in $\zeta(s+1) + \zeta(s)$ is $2/1^s = 2$. This is where the extra "$+1$" is accounted for.

Finally, the expression $$ \sum_{n \geq 1} \frac{1}{n^s} \sum_{\substack{d \mid n \\ 1 < d < n}} \frac{1}{d}$$ is equivalent to your conjectured description.

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  • $\begingroup$ Thank you! Is there a good resource I can refer to, to find such identities? Also, has this particular identity (or a form of it) been used/studied anywhere? $\endgroup$ Jun 30, 2020 at 18:09
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    $\begingroup$ I don't know of a good resource for this sort of identity, but I can recommend either Apostol's Intro to Analytic NT (relatively elementary) or Montgomery and Vaughan's Multiplicative NT (relatively sophisticated introduction) for classical introductions that give skills for this sort of thinking. There are lots of clever ways to use Dirichlet series, but I don't know of anything with these series or obviously related series here. $\endgroup$
    – davidlowryduda
    Jun 30, 2020 at 18:31

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