I was toying around with the Riemann Zeta function recently and noticed that I could get to a particular representation (valid for $Re(s)>1$) in a couple of different odd ways.

The first was by looking at perfect powers and non-perfect powers (by which I mean positive integers that are $>1$ and not equal to any perfect power). Here perfect powers are defined to be any integers of the form $a^b$ where $a,b\in \mathbb Z$ and $a,b>2$ (as a consequence, $1$ is neither perfect nor non-perfect).

It is clear that any positive integer $>1$ can be represented uniquely as some $a^k$ where $a$ is a non-perfect power and $k\in \mathbb Z^+$.

Using this fact, we can rewrite the traditional series representation of the Zeta function to get: $$\zeta(s)=\sum_{m=1}^{\infty}\frac{1}{m^s}=1+\sum_{r=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{a_r^{ns}}$$

where $a_r$ denotes the $r$th non-perfect number. One notes that the inner sum is a simple geometric sum- it may be evaluated to give $$\zeta(s)=1+\sum_{r=1}^{\infty}\bigg(\frac{1}{a_r^s-1}\bigg)$$

Now, we use the fact that each integer $>1$ is either a perfect power or a non-perfect power to give $$\zeta(s)=1+\sum_{m=2}^{\infty}\bigg(\frac{1}{m^s-1}\bigg)-\sum_{r=1}^{\infty}\bigg(\frac{1}{p_r^s-1}\bigg)$$

where $p_r$ denotes the $r$th perfect power. Finally, we note that every perfect power may be expressed uniquely as some $a^k$ where $a$ is a non-perfect power and $k$ is an integer $\geq 2$. Using this gives us $$\zeta(s)=1+\sum_{m=2}^{\infty}\bigg(\frac{1}{m^s-1}\bigg)-\sum_{n=2}^{\infty}\sum_{r=1}^{\infty}\bigg(\frac{1}{a_r^{ns}-1}\bigg)$$

But now we can apply the same tricks again on the inner sum, i.e. $$\zeta(s)=1+\sum_{m=2}^{\infty}\bigg(\frac{1}{m^s-1}\bigg)-\sum_{n=2}^{\infty}\Bigg(\sum_{m=2}^{\infty}\bigg(\frac{1}{m^{ns}-1}\bigg)-\sum_{r=1}^{\infty}\bigg(\frac{1}{p_r^{ns}-1}\bigg)\Bigg)\\=1+\sum_{m=2}^{\infty}\bigg(\frac{1}{m^s-1}\bigg)-\sum_{n=2}^{\infty}\sum_{m=2}^{\infty}\bigg(\frac{1}{m^{ns}-1}\bigg)+\sum_{n=2}^{\infty}\sum_{r=1}^{\infty}\bigg(\frac{1}{p_r^{ns}-1}\bigg)\\=1+\sum_{m=2}^{\infty}\bigg(\frac{1}{m^s-1}\bigg)-\sum_{n=2}^{\infty}\sum_{m=2}^{\infty}\bigg(\frac{1}{m^{ns}-1}\bigg)+\sum_{n_1=2}^{\infty}\sum_{n_2=2}^{\infty}\sum_{r=1}^{\infty}\bigg(\frac{1}{a_r^{n_1n_2s}-1}\bigg)$$

We can apply these couple of tricks repeatedly to get $$\zeta(s)=1+\sum_{m=2}^{\infty}\bigg(\frac{1}{m^s-1}\bigg)-\sum_{n_1=2}^{\infty}\sum_{m=2}^{\infty}\bigg(\frac{1}{m^{n_1s}-1}\bigg)+\sum_{n_1=2}^{\infty}\sum_{n_2=2}^{\infty}\sum_{m=2}^{\infty}\bigg(\frac{1}{m^{n_1n_2s}-1}\bigg)-...$$

But, there's another way we can get to this expansion- by using the Mobius function.

Since $\sum_{d:d|n}\mu(d)=1$ if $n=1$ and is $0$ otherwise, for any function $f$, we have $$f(s)=\sum_{i=1}^{\infty}\bigg(f(is)\sum_{d:d|i}\mu(d)\bigg)$$

If we rearrange the sum by collecting all the like $\mu(d)$ terms we get


When we let $f(s)=\zeta(s)-1$, we get $$\zeta(s)-1=\sum_{d=1}^{\infty}\sum_{j=1}^{\infty}\bigg(\mu(d)(\zeta(djs)-1)\bigg)=\sum_{d=1}^{\infty}\mu(d)\sum_{j=1}^{\infty}\sum_{m=2}^{\infty}\frac{1}{m^{djs}}=\sum_{d=1}^{\infty}\mu(d)\sum_{m=2}^{\infty}\frac{1}{m^{ds}-1}$$

Then, using this expansion of the Mobius function as the sum of the number of integer points on an n-dimensional hyperboloid and adding $1$ to both sides of the equation the same series may be obtained: $$\zeta(s)=1+\sum_{m=2}^{\infty}\bigg(\frac{1}{m^s-1}\bigg)-\sum_{n_1=2}^{\infty}\sum_{m=2}^{\infty}\bigg(\frac{1}{m^{n_1s}-1}\bigg)+\sum_{n_1=2}^{\infty}\sum_{n_2=2}^{\infty}\sum_{m=2}^{\infty}\bigg(\frac{1}{m^{n_1n_2s}-1}\bigg)-...$$

My question is, why do these different approaches (one, involving the Mobius function and, the other, using the basic properties of perfect powers) converge to the same result? Is there any way to remove the $\zeta$ function from all of this and get a clear-cut relationship between the Mobius function and the perfect/non-perfect numbers (in particular, could examining the error term of the expansion yield any results) or is this equality of approaches fundamentally dependent on the $\zeta$ function's properties? Or perhaps both approaches distract from a simpler possible derivation?

(please comment or edit for any corrections)

  • $\begingroup$ I voted this up as you've clearly thought about it and made a good research effort. $\endgroup$ – Martin Hansen Mar 10 at 19:09
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    $\begingroup$ It is not specific to $\zeta(s)$ : $\sum_{k=1}^\infty \frac{a(k)}{k^s-1}= \sum_{k=1}^\infty a(k) \sum_{m\ge 1} k^{-sm} = \sum_{n=1}^\infty n^{-s} \sum_{r=1}^{l(n)} a(n^{r/(l(n))})$ where $l(n)$ is the largest integer such that $n^{1/l(n)}$ is an integer, $l(1) = 1$. To obtain $\sum_{r=1}^{l(n)} a(n^{r/(ln)}) = 1$ it suffices to take $a(n) = 1$ if $l(n)=1$, $a(n) = 0$ otherwise. For an arbitrary $b(n)$ to obtain $\sum_{r=1}^{l(n)} a(n^{r/(ln)}) = b(n)$ for each $l(n)=1$ set $a(n) = b(n), m \ge 2,a(n^m) = b(n^m)-b(n^{m-1})$. $\endgroup$ – reuns Mar 10 at 19:12
  • $\begingroup$ @MartinHansen Thank you. I appreciate the feedback. Are there any ideas you have about the question? $\endgroup$ – Cardioid_Ass_22 Mar 10 at 19:23
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    $\begingroup$ @reuns Yes, I see-the representation of $\sum_{n=1}^{\infty} (b(n)n^{-s})$ as some $\sum_{k=1}^{\infty}\frac{a(k)}{k^s-1}$ is not unique to the case of $b(n)=1$ (for all $n$). But how does this relate to the larger expansion of $\zeta(s)$ being mentioned that is an expansion of $\zeta(s)$ as a series of series? And, even there, can the arbitrary Dirichlet series be 'removed' to find a direct relation between the Mobius function and perfect powers? $\endgroup$ – Cardioid_Ass_22 Mar 10 at 19:29
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    $\begingroup$ The representation is unique. About the second part of your post it holds for every analytic function such that the series converges absolutely. ${}{}{}{}$ $\endgroup$ – reuns Mar 10 at 19:39

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