1. ## Liouville's Function

Let

$\displaystyle f(n)=\underset{d|n}{\sum}d^{2}\lambda(d)$

where $\displaystyle \lambda(n)$ is Liouville's function.

How do I evaluate the sum

$\displaystyle \underset{n=1}{\overset{\infty}{\sum}}f(n)\: n^{-4}$

2. Originally Posted by Cairo
Let

$\displaystyle f(n)=\underset{d|n}{\sum}d^{2}\lambda(d)$

where $\displaystyle \lambda(n)$ is Liouville's function.

How do I evaluate the sum

$\displaystyle \underset{n=1}{\overset{\infty}{\sum}}f(n)\: n^{-4}$

Here's a broad hint: First, do you know the identity

$\displaystyle \frac{\zeta(2s)}{\zeta(s)} = \underset{n=1}{\overset{\infty}{\sum}}\frac{\lambd a(n)}{n^s}$

? It may be found in the Wikipedia article on Liouville's function, or as Theorem 300 in Hardy and Wright.

$\displaystyle \underset{n=1}{\overset{\infty}{\sum}}\underset{d| n}{\sum}\frac{d^{2}\lambda(d)}{n^4}$

Write out the first nine or so terms and collect together terms with $\displaystyle \lambda(1)$, $\displaystyle \lambda(2)$, $\displaystyle \lambda(3)$ and so on. Simplify, using the identity

$\displaystyle \zeta(4) = \underset{n=1}{\overset{\infty}{\sum}}\frac{1}{n^4 }$

where $\displaystyle \zeta$ is the Riemann zeta function. You should end up with an expression that involves the identity that I mentioned at the beginning of this post (with s = 2). Finally, use the well-known identities

$\displaystyle \zeta(2) = \frac{\pi^2}{6}$

and

$\displaystyle \zeta(4) = \frac{\pi^4}{90}$

to arrive at the final answer.

3. offtopic but can you tell me the book by hardy and right which you are suggesting here ?

4. Originally Posted by kira
offtopic but can you tell me the book by hardy and right which you are suggesting here ?
An Introduction to the Theory of Numbers by Hardy and Wright.

5. Thanks for your hint, Petek.

I'm not sure I follow

You should end up with an expression that involves the identity that I mentioned at the beginning of this post (with s = 2).
When I write out the terms of the sum, I can see that this is equal to $\displaystyle \zeta(4)$, but can't see how to evaluate it from here.

Am I right in thinking that

$\displaystyle \frac{\zeta(2s)}{\zeta(s)}=\underset{n=1}{\overset {\infty}{\sum}}\frac{\lambda(n)}{n^{s}}=\zeta(4)$

so that when s=2

$\displaystyle \underset{n=1}{\overset{\infty}{\sum}}\frac{\lambd a(n)}{n^{4}}=\zeta(2)\zeta(4)=\frac{\pi^{6}}{540}$

Regards

6. Here's what I had in mind:

$\displaystyle \underset{n=1}{\overset{\infty}{\sum}}\frac{f(n)}{ n^4}$
= $\displaystyle f(1) + \frac{f(2)}{2^4} + \frac{f(3)}{3^4} + \frac{f(4)}{4^4} + \cdots$
= $\displaystyle 1 + \frac{1 + 2^2\lambda(2)}{2^4} + \frac{1 + 3^2\lambda(3)}{3^4} + \frac{1 + 2^2\lambda(2) + 4^2\lambda(4)}{4^4} + \cdots$
=$\displaystyle (1 + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots)$
+$\displaystyle (\frac{2^2}{2^4} + \frac{2^2}{4^4} + \cdots)\lambda(2)$
+$\displaystyle (\frac{3^2}{3^4} + \frac{3^2}{6^4} + \cdots)\lambda(3)$
+$\displaystyle (\frac{4^2}{4^4} + \frac{4^2}{8^4} + \cdots)\lambda(4)$
+ ...

Can you take it from there? If not, what did you try?

7. I wrote the same and concluded that the sum is equal to zeta(4). I then used the identity you gave to reach the answer I provided above.

8. OK, now I see where we're not connecting. In one of your above posts you wrote

$\displaystyle \frac{\zeta(2s)}{\zeta(s)}=\underset{n=1}{\overset {\infty}{\sum}}\frac{\lambda(n)}{n^{s}}=\zeta(4)$
However, for s = 2, we get $\displaystyle \frac{\zeta(4)}{\zeta(2)}$, not $\displaystyle \zeta(4)$. Making that change should give you the same answer that I got.

9. Originally Posted by Petek
OK, now I see where we're not connecting. In one of your above posts you wrote

However, for s = 2, we get $\displaystyle \frac{\zeta(4)}{\zeta(2)}$, not $\displaystyle \zeta4)$. Making that change should give you the same answer that I got.
I'm not sure I'm following you here.

10. First, remember that we're trying to evaluate

$\displaystyle \underset{n=1}{\overset{\infty}{\sum}}f(n)\: n^{-4}$

Refer to my equations in post #6 and let's continue from there:

$\displaystyle \underset{n=1}{\overset{\infty}{\sum}}\frac{f(n)}{ n^4}$

$\displaystyle = \zeta(4) \frac{\lambda(1)}{1^2}$

$\displaystyle + \frac{2^2}{2^4}(1 + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots)\lambda(2)$

$\displaystyle + \frac{3^2}{3^4}(1 + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots)\lambda(3)$

$\displaystyle + \frac{4^2}{4^4}(1 + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots)\lambda(4)$

$\displaystyle + \cdots$

$\displaystyle = \zeta(4)\frac{\lambda(1)}{1^2} + \zeta(4)\frac{\lambda(2)}{2^2} + \zeta(4) \frac{\lambda(3)}{3^2} + \cdots$

$\displaystyle = \zeta(4) \underset{n=1}{\overset{\infty}{\sum}}\frac{\lambd a(n)}{n^{2}}$

(Since $\displaystyle \frac{2^2}{2^4} = \frac{1}{2^2},\ \frac{3^2}{3^4} = \frac{1}{3^2}$, and so on.)

$\displaystyle =\frac{\zeta^2(4)}{\zeta(2)}$

11. pi^6 / 1350 ?

Thanks

12. Originally Posted by Cairo
pi^6 / 1350 ?

Thanks
That's what I got. You're welcome.