Identity involving a relation between $\zeta(s)$ and $\zeta(s+1)$ for integers s > 1 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.
 A: 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.
