Why does the $\sum_{n>1}(\zeta(n)-1)=1?$

Question

While I was looking at the values of the zeta function for the first natural numbers, I noticed that the sum of the values minus $1$, converge to $1$. Better put: $$\sum_{n=2}^{\infty} \left(\zeta(n)-1\right) = 1 $$ Furthermore, if you use only the even numbers for the zeta function, the sum will converge to $\frac{3}{4}$, or $$\sum_{n=1}^{\infty} \left(\zeta(2n)-1\right) = \frac{3}{4}$$

Leaving $$\sum_{n=2}^{\infty} \left(\zeta(2n-1)-1 \right)= \frac{1}{4}$$

This is probably common knowledge among mathematicians, but I couldn't find much about it on the internet. Is there a proof of this or perhaps even a simple explanation why this is so?

Answer 1

Note: $$\begin{align}\sum_{n=2}^\infty (\zeta(n)-1) &= \sum_{n=2}^\infty \sum_{k=2}^\infty\frac{1}{k^n}\\ &=\sum_{k=2}^\infty\sum_{n=2}^\infty \frac{1}{k^n} \end{align}$$

And $$\sum_{n=2}^{\infty} \frac{1}{k^n} =\frac{1}{k^2}\frac{1}{1-\frac{1}{k}}= \frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}.$$

For $\zeta(2n)$ case:

$$ \sum_{n=1}^{\infty} \frac{1}{k^{2n}} = \frac{1}{k^2-1} = \frac{1}{2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right)$$

More generally, if $f(z)=\sum_{n=2}^\infty a_nz^n$ has radius of convergence more than $\frac{1}2$, then:

$$\sum_{n=2}^\infty a_n(\zeta(n)-1) = \sum_{k=2}^\infty f\left(\frac1k\right)$$

This can be used to show that $$\sum_{n=2}^\infty \frac{\zeta(n)-1}{n} = 1-\gamma$$ where $\gamma$ is the Euler–Mascheroni constant. Using the standard limit for $\gamma$, we see that:

$$\lim_{N\to\infty} \left(\log N -\sum_{n=2}^N \frac{\zeta(n)}{n}\right) = 0$$

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Very late comment

I just noticed that if $f(z)=\sum_{n=2}^\infty a_nz^n$ has radius of convergence greater than $1,$ we get:

$$\sum_{n=2}^\infty a_n \zeta(n) = \sum_{k=1}^\infty f\left(\frac 1k\right)$$

Answer 2

Using the series representation of the Riemann-Zeta function

$$\zeta(n)=\sum_{k=1}^{\infty}\frac{1}{k^n}$$

gives

$$\begin{align} \sum_{n=2}^{\infty}\left(\sum_{k=1}^{\infty}\frac{1}{k^n}-1\right)&=\sum_{k=2}^{\infty}\sum_{n=2}^{\infty}\frac{1}{k^n}\\\\ &=\sum_{k=2}^{\infty}\left(\frac{1/k^2}{1-1/k}\right)\\\\ &=\sum_{k=2}^{\infty}\left(\frac{1}{k-1}-\frac{1}{k}\right)\\\\ &=1 \end{align}$$

Answer 3

There is a whole series of formulae like this. The proofs are all along the lines of writing out the $\zeta$ sums, changing the order of summation (making sure that this is valid, of course!), and doing the interior sum. In this case, the double sum will be $$ \begin{align} \sum_{n=2}^{\infty} (\zeta(n)-1) &= \sum_{n=2}^{\infty} \sum_{k=2}^{\infty} \frac{1}{k^n} \\ &= \sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{k^n} \\ &= \sum_{k=2}^{\infty} \frac{1}{k^2(1-1/k)} \\ &= \sum_{k=2}^{\infty} \frac{1}{k(k-1)} \\ &= \sum_{k=2}^{\infty} \left( \frac{1}{k} - \frac{1}{k-1} \right), \end{align} $$ which it is easy to see telescopes down to $1$. Some other examples can be found here.

Answer 4

For any $s>1$ we have $$\zeta(s)-1=\int_{0}^{+\infty}\frac{x^{s-1}}{(s-1)!}\cdot\frac{dx}{e^x(e^x-1)}\tag{1}$$ hence: $$ \sum_{n\geq 2}\left(\zeta(n)-1\right) = \int_{0}^{+\infty}\frac{e^x-1}{e^x(e^x-1)}\,dx = 1 \tag{2} $$ and: $$ \sum_{n\geq 2}\left(\zeta(2n-1)-1\right) = \int_{0}^{+\infty}\frac{\cosh(x)-1}{e^x(e^x-1)}\,dx=\int_{0}^{+\infty}\frac{e^{-x}-e^{-2x}}{2}\,dx=\frac{1}{4}\tag{3} $$ $$ \sum_{n\geq 1}\left(\zeta(2n)-1\right) = \int_{0}^{+\infty}\frac{\sinh(x)}{e^x(e^x-1)}\,dx=\int_{0}^{+\infty}\frac{e^{-x}+e^{-2x}}{2}\,dx=\frac{3}{4}.\tag{4} $$

Answer 5

Simple laymans proof: Start with the Harmonic series

1. $1 +1/2 +1/3 + 1/4 + 1/5 $ ...etc

2. you can sum the powers of every individual Integer independently i.e.

3. $ 1/2^1 +1/2^2 +1/2^3 $ .... sums to 1 (2-1)

4. $ 1/3^1 +1/3^2 +1/3^3 $ .... sums to 1/2 (3-1)

5. $ 1/5^1 + 1/5^2 +1/5^3 $ ... sums to 1/4 (5-1)

6. As a rule $$\sum_{k=1}^\infty 1/N^K = 1/(N-1)$$

7. We can now rearrange the original series as $1 +1 + 1/2 +1/4 +1/5 +1/6 + 1/9 ....$

8. This results in an additional 1 in the beginng of the series while many terms are missing.

9. The missing terms are 1 less than any power of any integer. Since these numbers cannot be used to form new series as they have already been used in the series beginning with the base number.

10. The missing terms converge to 1. The missing terms correspond to all values of the zeta function. It follows that $$\sum_{n=2}^\infty (\zeta(n)-1)= 1$$

Amateur proof