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Is there a simple way to prove the following identity

$$\sum_{s=2}^{\infty}(\zeta(s)-1)=1$$

I imagine that there is some nice thing one can do using the Euler product representation, though it currently escapes me.

EDIT:

I can show that this is equal to

$$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\dots=\sum_{n=1}^\infty\frac{1}{n^2+n}$$

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$$\sum\limits_{s=2}^{\infty} (\zeta(s)-1) = \sum\limits_{s=2}^{\infty} \sum\limits_{k=2}^{\infty} \frac{1}{k^s} = \sum\limits_{k=2}^{\infty} \sum\limits_{s=2}^{\infty} \frac{1}{k^s} = \sum\limits_{k=2}^{\infty} \frac{1}{k(k-1)}$$

$$= \sum\limits_{k=2}^{\infty} \frac{1}{k-1} - \frac{1}{k} $$

This last sum is telescoping.

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    $\begingroup$ Thanks, I just figured this out and was about to post it! $\endgroup$ – ruadath Nov 23 '16 at 4:45
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    $\begingroup$ I always figure things out right after I post them on stackexchange $\endgroup$ – D_S Nov 23 '16 at 4:46

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