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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?

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5 Answers 5

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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:

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


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:

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

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    $\begingroup$ You're extraordinarily fast! $\endgroup$
    – Mark Viola
    Sep 13, 2015 at 17:17
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Using the series representation of the Riemann-Zeta function

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

gives

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

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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.

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  • $\begingroup$ I understand why we need to be careful about changing the orders of infinite summations, but why is it justifiable in this particular case? Is there some powerful lemma or something that can be used to get around this issue quickly for this example? $\endgroup$
    – Patch
    Jun 23, 2022 at 1:07
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    $\begingroup$ @Patch If the answer is still relevant, this is just a particular case of Fubini's theorem or (more easily seen to be applicable in this case) Tonelli's theorem further down on the same page. How is this related to integration? For this, see the counting measure and recall the definition of the Lebesgue integral - if you had a measure theory course already. As a general rule, this trick always works if the integrand is non-negative over all attained values of the integral (or sum). $\endgroup$
    – TheOutZ
    Sep 18, 2022 at 17:22
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    $\begingroup$ Replace integrand with summand and integral with summing variables in my comment above. Measurability can, in general, only be ignored in case of countable spaces like $\mathbb{N}$ since one can show that every of those functions mapping to $\mathbb{R}$ is measurable. Hence we only need non-negativity in the case of arbitarily many nested infinite sums. $\endgroup$
    – TheOutZ
    Sep 18, 2022 at 17:30
  • $\begingroup$ Thank you for that! $\endgroup$
    – Patch
    Sep 19, 2022 at 3:52
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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: $$ \sum_{n\geq 2}\left(\zeta(2n-1)-1\right) = \int_{0}^{+\infty}\frac{\cosh(x)-1}{e^x(e^x-1)}\,dx=\int_{0}^{+\infty}\frac{e^{-x}-e^{-2x}}{2}\,dx=\frac{1}{4}\tag{3} $$ $$ \sum_{n\geq 1}\left(\zeta(2n)-1\right) = \int_{0}^{+\infty}\frac{\sinh(x)}{e^x(e^x-1)}\,dx=\int_{0}^{+\infty}\frac{e^{-x}+e^{-2x}}{2}\,dx=\frac{3}{4}.\tag{4} $$

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    $\begingroup$ +1. Very helpful. Likewise we have, $$ \frac{\zeta(s)-1}{s}=\int_{0}^{\infty}\frac{x^{s}}{s!}\cdot\frac{dx}{ x e^x(e^x-1)}$$ $$ \sum_{s=2}^\infty\frac{\zeta(s)-1}{s}= \int_0^\infty \frac{e^x-x-1}{xe^x(e^x-1)}dx$$ $$=\int_{0}^{\infty}\frac{dx}{e^{x}}-\int_{0}^{\infty}\left(\frac{1}{e^{x}-1}-\frac{1}{xe^{x}}\right)dx$$ $$1-\gamma$$ $\endgroup$
    – Mason
    Apr 23, 2020 at 2:53
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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

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