Inverse of the Riemann Zeta Function

[DeadOriginal]

Homework Statement

I wish to prove that for s>1
$$
\sum\limits_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\prod_{p}(1-p^{-s})=\frac{1}{\zeta(s)}.
$$

The Attempt at a Solution

(1) I first showed that
$$
\prod_{p}(1-p^{-s})=\frac{1}{\zeta(s)}.
$$
It was a given theorem in the text that
$$
\zeta(s)=\prod_{p}\frac{1}{1-p^{-s}}
$$
so I said
$$
\frac{1}{\zeta(s)}=\frac{1}{\prod_{p}\frac{1}{1-p^{-s}}}=\frac{\prod_{p}1}{\prod_{p}\frac{1}{1-p^{-s}}}=\prod_{p}\frac{1}{\frac{1}{1-p^{-s}}}=\prod_{p}(1-p^{-s}).
$$
I am worried about this because $1^{\infty}$ isn't exactly defined.

(2) I then expanded out the product $\prod_{p}(1-p^{-s})$ and saw that the general term in the series that I got was in the form $\frac{\mu(n)}{n^{s}}$ where $n=p_{1}^{e_{1}}\cdots p_{k}^{e_{k}}$ for $e_{i}=0$ or $e_{i}=1$. Since $p_{i}^{0}=1$ I can remove all of the terms that are equal to 1 to obtain $n=p_{1}\cdots p_{j}$ and then by another theorem that was given to me I have $\mu(n)=(-1)^{j}$. I noticed that in the expansion of the product, the fractions that had a even number of primes in the denominator were positive and the ones with an odd number of primes were negative so that would give me the general term $\frac{\mu(n)}{n^{s}}$. Thus
$$
\sum\limits_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\prod_{p}(1-p^{-s}).
$$
I am worried about this argument because it doesn't seem to be rigorous to me. What do you think?

 

[Dick]

Homework Statement

I wish to prove that
$$
\sum\limits_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\prod_{p}(1-p^{-s})=\frac{1}{\zeta(s)}.
$$

The Attempt at a Solution

(1) I first showed that
$$
\prod_{p}(1-p^{-s})=\frac{1}{\zeta(s)}.
$$
It was a given theorem in the text that
$$
\zeta(s)=\prod_{p}\frac{1}{1-p^{-s}}
$$
so I said
$$
\frac{1}{\zeta(s)}=\frac{1}{\prod_{p}\frac{1}{1-p^{-s}}}=\frac{\prod_{p}1}{\prod_{p}\frac{1}{1-p^{-s}}}=\prod_{p}\frac{1}{\frac{1}{1-p^{-s}}}=\prod_{p}(1-p^{-s}).
$$
I am worried about this because $1^{\infty}$ isn't exactly defined.

(2) I then expanded out the product $\prod_{p}(1-p^{-s})$ and saw that the general term in the series that I got was in the form $\frac{\mu(n)}{n^{s}}$ where $n=p_{1}^{e_{1}}\cdots p_{k}^{e_{k}}$ for $e_{i}=0$ or $e_{i}=1$. Since $p_{i}^{0}=1$ I can remove all of the terms that are equal to 1 to obtain $n=p_{1}\cdots p_{j}$ and then by another theorem that was given to me I have $\mu(n)=(-1)^{j}$. I noticed that in the expansion of the product, the fractions that had a even number of primes in the denominator were positive and the ones with an odd number of primes were negative so that would give me the general term $\frac{\mu(n)}{n^{s}}$. Thus
$$
\sum\limits_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\prod_{p}(1-p^{-s}).
$$
I am worried about this argument because it doesn't seem to be rigorous to me. What do you think?

I think it's pretty much the usual argument. $1^{\infty}$ is exactly defined. It's 1. You might be confusing this with lim(x^y) where x->1 and y->infinity. That's undefined since it depends on the exact behavior of x and y.

 

[DeadOriginal]

Hmm. The way the proof goes in my text for $\zeta(s)=\prod_{p}\left(\frac{1}{1-p^{-s}}\right)$ is that a finite product, $P_{k}(s)$ is defined. The author then uses the fact that the finite product is equal to a finite sum with the general term $\frac{1}{n^{s}}$. The author then goes to show that $|P_{k}(s)-\zeta(s)|<\epsilon$. I tried to do this but I kept getting the feeling that something was wrong with my proof.

I basically defined $P_{k}(s)=\sum\limits_{n\in A_{k}}\frac{\mu(n)}{n^{s}}$ where $A_{k}=\{n:n=p_{1}^{e_{1}}\cdots p_{k}^{e_{k}}, e_{i}=0,1\}$ after seeing the general pattern that I described in my original post in part (2). I also noted that each $n\not\in A_{k}$ is divisible by some prime $p>p_{k}$. Then I said since s>1
$$
|P_{k}(s)-\frac{1}{\zeta(s)}|\leq\sum\limits_{n\not\in A_{k}}\frac{\mu(n)}{n^{s}}\leq\sum\limits_{n>p_{k}}\frac{|\mu(n)|}{n^{s}}\leq\sum\limits_{n>p_{k}}\frac{1}{n^{s}}=\zeta(s)-\sum\limits_{n\leq p_{k}}\frac{1}{n^{s}}.
$$
Since s>1, the partial sums of the series $\sum\frac{1}{n^{s}}$ converges to $\zeta(s)$ so then
$$
\sum\limits_{n\leq p_{k}}\frac{1}{n^{s}}\rightarrow\zeta(s)
$$
as $k\rightarrow\infty$. This shows that $|P_{k}(s)-\frac{1}{\zeta(s)}|\rightarrow 0$ as $k\rightarrow 0$.

I am not sure why but I feel like something is wrong there. In particular, the part where I said $\sum\limits_{n>p_{k}}\frac{|\mu(n)|}{n^{s}}\leq\sum\limits_{n>p_{k}} \frac{1}{n^{s}}=\zeta(s)-\sum\limits_{n\leq p_{k}}\frac{1}{n^{s}}$ makes me feel a little.. uneasy with the proof for some reason. $\zeta(s)-\sum\limits_{n\leq p_{k}}\frac{1}{n^{s}}$ was the same exact bound that was found for $|P_{k}(s)-\zeta(s)|$. Could it be possible that the same bound works for $|P_{k}(s)-\frac{1}{\zeta(s)}|$?

 

[Dick]

Hmm. The way the proof goes in my text for $\zeta(s)=\prod_{p}\left(\frac{1}{1-p^{-s}}\right)$ is that a finite product, $P_{k}(s)$ is defined. The author then uses the fact that the finite product is equal to a finite sum with the general term $\frac{1}{n^{s}}$. The author then goes to show that $|P_{k}(s)-\zeta(s)|<\epsilon$. I tried to do this but I kept getting the feeling that something was wrong with my proof.

I basically defined $P_{k}(s)=\sum\limits_{n\in A_{k}}\frac{\mu(n)}{n^{s}}$ where $A_{k}=\{n:n=p_{1}^{e_{1}}\cdots p_{k}^{e_{k}}, e_{i}=0,1\}$ after seeing the general pattern that I described in my original post in part (2). I also noted that each $n\not\in A_{k}$ is divisible by some prime $p>p_{k}$. Then I said since s>1
$$
|P_{k}(s)-\frac{1}{\zeta(s)}|\leq\sum\limits_{n\not\in A_{k}}\frac{\mu(n)}{n^{s}}\leq\sum\limits_{n>p_{k}}\frac{|\mu(n)|}{n^{s}}\leq\sum\limits_{n>p_{k}}\frac{1}{n^{s}}=\zeta(s)-\sum\limits_{n\leq p_{k}}\frac{1}{n^{s}}.
$$
Since s>1, the partial sums of the series $\sum\frac{1}{n^{s}}$ converges to $\zeta(s)$ so then
$$
\sum\limits_{n\leq p_{k}}\frac{1}{n^{s}}\rightarrow\zeta(s)
$$
as $k\rightarrow\infty$. This shows that $|P_{k}(s)-\frac{1}{\zeta(s)}|\rightarrow 0$ as $k\rightarrow 0$.

I am not sure why but I feel like something is wrong there. In particular, the part where I said $\sum\limits_{n>p_{k}}\frac{|\mu(n)|}{n^{s}}\leq\sum\limits_{n>p_{k}} \frac{1}{n^{s}}=\zeta(s)-\sum\limits_{n\leq p_{k}}\frac{1}{n^{s}}$ makes me feel a little.. uneasy with the proof for some reason. $\zeta(s)-\sum\limits_{n\leq p_{k}}\frac{1}{n^{s}}$ was the same exact bound that was found for $|P_{k}(s)-\zeta(s)|$. Could it be possible that the same bound works for $|P_{k}(s)-\frac{1}{\zeta(s)}|$?

I guess I don't see anything in there that makes me uneasy. You are just trying to derive the proof of $\frac{1}{\zeta(s)}$ from the given proof of $\zeta(s)$, right?

 

[DeadOriginal]

I guess I don't see anything in there that makes me uneasy. You are just trying to derive the proof of $\frac{1}{\zeta(s)}$ from the given proof of $\zeta(s)$, right?

Yea. Maybe it's because I don't deal with infinite products often and need more experience with them.

 

[Dick]

Yea. Maybe it's because I don't deal with infinite products often and need more experience with them.

Maybe, but you can express all of the infinite products in terms into finite partial products and then into finite partial sums and then taken the limit. Seems ok to me.

 

[DeadOriginal]

Alright. Thanks!

 

[jackmell]

That's not the inverse of zeta, rather it's only the reciprocal. But surprisingly, we can construct the actual inverse of the zeta function. That is, if $w(s)=\zeta(s)$, then what is $s(w)$.

Anyway, I think the title of this thread should reflect this distinction.

 

[DeadOriginal]

That's not the inverse of zeta, rather it's only the reciprocal. But surprisingly, we can construct the actual inverse of the zeta function. That is, if $w(s)=\zeta(s)$, then what is $s(w)$.

Anyway, I think the title of this thread should reflect this distinction.

Yea sorry about that. When I realized my mistake, it was already too late to change it.