# Thread: Prove some identities! (Round two)

1. ## Prove some identities! (Round two)

And you thought it was over?

Problem: Compute

$\displaystyle \sum_{n=2}^{\infty}\left(\zeta(n)-1\right)$

,

$\displaystyle \sum_{n=1}^{\infty}\left(\zeta(2n)-1\right)$

and,

$\displaystyle \sum_{n=2}^{\infty}\left(\zeta(2n-1)-1\right)$

2. Cool ones!

$\displaystyle{
\sum_{n=2}^\infty (\zeta(n)-1) = \sum_{n=2}^\infty\sum_{m=2}^\infty \frac{1}{m^n} = \sum_{m=2}^\infty\sum_{n=2}^\infty \frac{1}{m^n}
}$

$\displaystyle{=
\sum_{m=2}^\infty \frac{1}{m^2-m} = \sum_{m=2}^\infty \left(\frac{1}{m-1}-\frac{1}{m}\right) = 1
}$

Likewise:

$\displaystyle{
\sum_{n=1}^\infty (\zeta(2n)-1) = \sum_{m=2}^\infty\sum_{n=1}^\infty \frac{1}{m^{2n}} = \sum_{m=2}^\infty\frac{1}{m^2-1}
}$

$\displaystyle{
= \frac{1}{2}\sum_{m=2}^\infty \left(\frac{1}{m-1}-\frac{1}{m+1}\right) = \frac{3}{4}
}$

And the last one is the difference between the first and the second, that is:

$\displaystyle{
\sum_{n=1}^\infty (\zeta(2n+1)-1) = \frac{1}{4}
}$

3. Originally Posted by Unbeatable0
Cool ones!

$\displaystyle{
\sum_{n=2}^\infty (\zeta(n)-1) = \sum_{n=2}^\infty\sum_{m=2}^\infty \frac{1}{m^n} = \sum_{m=2}^\infty\sum_{n=2}^\infty \frac{1}{m^n}
}$

$\displaystyle{=
\sum_{m=2}^\infty \frac{1}{m^2-m} = \sum_{m=2}^\infty \left(\frac{1}{m-1}-\frac{1}{m}\right) = 1
}$

Likewise:

$\displaystyle{
\sum_{n=1}^\infty (\zeta(2n)-1) = \sum_{m=2}^\infty\sum_{n=1}^\infty \frac{1}{m^{2n}} = \sum_{m=2}^\infty\frac{1}{m^2-1}
}$

$\displaystyle{
= \frac{1}{2}\sum_{m=2}^\infty \left(\frac{1}{m-1}-\frac{1}{m+1}\right) = \frac{3}{4}
}$

And the last one is the different between the first and the second, that is:

$\displaystyle{
\sum_{n=1}^\infty (\zeta(2n+1)-1) = \frac{1}{4}
}$
Right, not too hard, but interesting indeed. Of course using the same methodology shows that $\displaystyle \sum_{n=1}^{\infty}(-1)^n\left(\zeta(n)-1\right) =\frac{1}{2}$. This one is slightly harder. Try

$\displaystyle \sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n}$

4. Originally Posted by Drexel28
Right, not too hard, but interesting indeed. Of course using the same methodology shows that $\displaystyle \sum_{n=1}^{\infty}(-1)^n\left(\zeta(n)-1\right) =\frac{1}{2}$. This one is slightly harder. Try

$\displaystyle \sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n}$
Of course, the anternating one is just the difference between the second and the last ones you proposed in the first post.

I find it very interesting that they're rational.

The other one you proposed is the same method:

$\displaystyle{
\sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n} = \sum_{m=2}^\infty\sum_{n=2}^\infty \frac{1}{nm^n} = \sum_{m=2}^\infty\left(\ln m - \ln (m-1) - \frac{1}{m}\right)
}$

$\displaystyle{
=1+\lim_{n\rightarrow\infty} \left(\ln(n)-\sum_{m=1}^n\frac{1}{m}\right) = 1-\gamma
}$

5. Originally Posted by Unbeatable0
Of course, the anternating one is just the difference between the second and the last ones you proposed in the first post.

I find it very interesting that they're rational.

The other one you proposed is the same method:

$\displaystyle{
\sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n} = \sum_{m=2}^\infty\sum_{n=2}^\infty \frac{1}{nm^n} = \sum_{m=2}^\infty\left(\ln m - \ln (m-1) - \frac{1}{m}\right)
}$

$\displaystyle{
=1+\lim_{n\rightarrow\infty} \left(\ln(n)-\sum_{m=1}^n\frac{1}{m}\right) = 1-\gamma
}$

$\displaystyle \sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(2n+1)}$

6. Originally Posted by Drexel28

$\displaystyle \sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(2n+1)}$
Spoiler:

We know $\displaystyle \frac{\sin\pi x}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right) \implies \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n}$.

Thus we see $\displaystyle \sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(2n+1)} = \int_0^1 \log(\pi x) dx - \int_0^1 \log(\sin\pi x) dx = \log\pi-1+\log2 = \log(2\pi)-1$.

7. Originally Posted by chiph588@
We know $\displaystyle \frac{\sin\pi x}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right) \implies \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n}$.

Thus we see $\displaystyle \sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(2n+1)} = \int_0^1 \log(\pi x) dx - \int_0^1 \log(\sin\pi x) dx = \log\pi-1+\log2 = \log(2\pi)-1$.

Right! Now someone else post a problem so I can solve one! haha

8. Here are some nice ones. Compute:

$\displaystyle
\sum_{n=1}^\infty \frac{|\mu (n)|}{n^s}$

$\displaystyle
\sum_{n=1}^\infty \frac{\sigma(n)}{n^s}$

$\displaystyle
\sum_{n=1}^\infty \frac{2^{\omega(n)}}{n^s}$

where $\mu(n)$ is the Mobius function, $\sigma(n)$ is the divisor function, and $\omega(n)$ counts the number of distinct primes dividing $n$.

Edit: just to clarify, by $\sigma(n)$ I was referring, as usually denoted by this letter, to the function returning the sum of the divisors of $n$, and not the number of divisors, which is also called the divisor function.

9. Originally Posted by Unbeatable0

$\displaystyle
\sum_{n=1}^\infty \frac{2^{\omega(n)}}{n^s}$

Note that we require $\text{Re}(s) > 1$. I will do one, because they all can be derived in a similar fashion.

Spoiler:

It's not too hard to see $2^\omega(n)$ is multiplicative. Hence

\begin{aligned} \displaystyle \sum_{n=1}^\infty \frac{2^{\omega(n)}}{n^s} &= \prod_p\sum_{k=0}^\infty \frac{2^\omega(p^k)}{p^{sk}}\\
&= \prod_p\left(1+2\sum_{k=1}^\infty\frac1{p^{sk}}\ri ght)\\
&= \prod_p\left(1+2\frac{p^{-s}}{1-p^{-s}}\right)\\
&= \prod_p \frac{1+p^{-s}}{1-p^{-s}}\\
&= \prod_p \frac{1-p^{-2s}}{(1-p^{-s})^2}\\
&= \frac{\zeta^2(s)}{\zeta(2s)} \end{aligned}

Try this one: Compute
$\displaystyle \int_0^1 (-\log x)^a dx$

10. What's required to switch the order of summation? Absolute convergence of the inner sum?

Originally Posted by Unbeatable0
Cool ones!

$\displaystyle{
\sum_{n=2}^\infty (\zeta(n)-1) = \sum_{n=2}^\infty\sum_{m=2}^\infty \frac{1}{m^n} = \sum_{m=2}^\infty\sum_{n=2}^\infty \frac{1}{m^n}
}$

$\displaystyle{=
\sum_{m=2}^\infty \frac{1}{m^2-m} = \sum_{m=2}^\infty \left(\frac{1}{m-1}-\frac{1}{m}\right) = 1
}$

Likewise:

$\displaystyle{
\sum_{n=1}^\infty (\zeta(2n)-1) = \sum_{m=2}^\infty\sum_{n=1}^\infty \frac{1}{m^{2n}} = \sum_{m=2}^\infty\frac{1}{m^2-1}
}$

$\displaystyle{
= \frac{1}{2}\sum_{m=2}^\infty \left(\frac{1}{m-1}-\frac{1}{m+1}\right) = \frac{3}{4}
}$

And the last one is the difference between the first and the second, that is:

$\displaystyle{
\sum_{n=1}^\infty (\zeta(2n+1)-1) = \frac{1}{4}
}$

11. $\displaystyle \sum_{n=1}^{\infty} \frac{\sigma(n) }{n^s }$

I think all of us would factorise the summation as this arithmetic function is multiplicative !

$\displaystyle = \prod_{\text{all primes p }} \left( 1 + \frac{\sigma(p)}{p^s} + \frac{\sigma(p^2)}{p^{2s}} + ... \right)$

$\displaystyle = \prod_{\text{all primes p }} \left[ \sum_{k=0}^{\infty} \left(\frac{ p^{k+1} - 1 }{p^{ks}(p - 1) } \right) \right]$

$\displaystyle = \prod_{\text{all primes p }} \frac{1}{p-1} \left( \frac{p}{1 - 1/p^{s-1} } - \frac{1}{1 - 1/p^s } \right)$

$\displaystyle = \prod_{\text{all primes p }} \frac{1}{p-1} \left( \frac{p-1}{ (1 - 1/p^{s-1} )(1 - 1/p^s) } \right)$

$\displaystyle = \prod_{\text{all primes p }} \frac{1}{(1 - 1/p^{s-1} )(1 - 1/p^s) }$

$\displaystyle = \zeta(s) \zeta(s-1)$

About the first problem , I remember it was first discovered by Ramanujan , he expanded the product :

$\displaystyle \frac{1}{\zeta(s)} = \prod_{\text{all primes p }} \left( 1 - \frac{1}{p^s} \right)$

As each factor provides only a prime but not a power of it , all the summands of the expansion should be square-free , and the signs of them exactly indicate the moebius function .

Therefore , $\displaystyle \frac{1}{\zeta(s) } = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}$

But this time , absolute value is taken :

$\displaystyle \sum_{n=1}^{\infty} \frac{|\mu (n)|}{n^s} = \prod_{\text{all primes p }} \left( 1 + \frac{1}{p^s} \right) = \frac{ \zeta(s)}{\zeta(2s)}$

12. $\int^{1}_{0} (-\ln x)^{a} \ dx$

$= (-1)^{a} \int^{1}_{0} (\ln x)^{a} \ dx$

let $x = e^{-u}$

$= (-1)^{a} \int_{\infty}^{0} (-u)^{a} (-e^{-u}) \ du$

$= \int^{\infty}_{0} u^{a} e^{-u} \ du$

$= \Gamma (a+1)$

13. Evaluate $\displaystyle \int_0^1 \int_0^1 \frac{\log(xy)}{xy-1} dxdy$.

14. Originally Posted by chiph588@
Evaluate $\displaystyle \int_0^1 \int_0^1 \frac{\log(xy)}{xy-1} dxdy$.
Damnit, I had to go and all the cool number theory ones were already done!

Spoiler:

[CENTER]

This particular case falls to pretty easy techniques

\displaystyle \begin{aligned}\int_0^1\int_0^1\frac{\ln(xy)}{1-xy}\text{ }dx\text{ }dy &= \int_0^1\int_0^1 \sum_{n=0}^{\infty}\ln^s(xy)(xy)^n\text{ }dx\text{ }dy\\ &=\sum_{n=0}^{\infty}\int_0^1\int_0^1 \ln(xy)(xy)^n\text{ }dx\text{ }dy\\ &= \sum_{n=0}^{\infty}\int_0^1 \frac{y^n\left((n+1)\log(y)-1\right)}{(n+1)^2}\text{ }dy\\ &= -\sum_{n=0}^{\infty}\frac{2}{(n+1)^3}\\ &= -2\zeta(3)\end{aligned}

But, it's slightly harder (not much, ask if you're interested) that

$\displaystyle \int_0^1\int_0^1\frac{\ln^s(xy)}{1-xy}\text{ }dx\text{ }dy=(-1)^s\Gamma(s+2)\zeta(s+2)\quad \text{Re}(s)>1$

15. Spoiler:
$\displaystyle \int_0^1 \int_0^1 \frac{\ln^s(xy)}{xy - 1 }~dx dy$

Sub. $xy = u$ ( $y$ being constant ) we have

$\displaystyle \int_0^1 \int_0^1 \frac{\ln^s(xy)}{xy - 1 }~dx~dy$

$\displaystyle = \int_0^1 \int_0^y \frac{\ln^s(u)}{u-1} \frac{du}{y} ~dy$

$\displaystyle = \int_0^1 \int_u^1 \frac{dy}{y}~ \frac{\ln^s(u)}{u-1}~du$

$\displaystyle = - \int_0^1 \frac{\ln^{s+1}(u)}{u-1} ~du$

Sub. $\displaystyle u = e^{-t}$

$\displaystyle = - \int_0^{\infty} \frac{(-1)^{s+1} t^{s+1} }{ 1 - e^t }~dt$

$\displaystyle = (-1)^{s+1} \int_0^{\infty} \frac{t^{s+1}}{e^t - 1 } ~dt$

$\displaystyle = (-1)^{s+1} \Gamma(s+2) \zeta(s+2)$

$\displaystyle \int_0^1 \int_0^1 \int_0^1 \frac{dxdydz}{(1 + x^2 + y^2 + z^2)^2 }$ and

$\displaystyle M(x,y) = \frac{\pi}{4} ~ \frac{x+y}{ \big K\left[ \left( \frac{x-y}{x+y} \right)^2 \right] }$

where $\displaystyle K(m) = \int_0^{\pi/2} \frac{dx}{\sqrt{ 1 - m \sin^2(x) } }$ and $M(x,y)$ is arithmetic geometric mean :

$\displaystyle a_1 = \frac{x+y}{2} ~,~ g_1 = \sqrt{xy}$ Then

$\displaystyle a_{n+1} = \frac{a_n + g_n }{2} ~,~ g_{n+1} = \sqrt{a_n g_n }$

$\displaystyle M(x,y) = \lim_{n\to \infty} a_n = \lim_{n\to \infty} g_n$

EDIT : I have made a very big mistake on the first problem , the integrand should be squared !

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