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How does one prove the zeta function identity

$$\sum_{s=2}^{\infty}\left(1-\sum_{n=1}^{\infty}\frac{1}{n^s}\right)=-1 \;?$$

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We have to deal with the problem of whether interchanging the order of summation is legitimate (sometimes it isn't). – André Nicolas Feb 9 '12 at 16:40

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

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

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