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If you consider the series of reciprocal positive integers (the harmonic series), the sum diverges. However, some subsets of this series, such as the reciprocal cubes, will sum to something finite. I was interested in the kind of "crossover", where I add more terms to the sum and it becomes divergent, so I was wondering about summing all the terms like this.

Take the sum over all the reciprocal squares, cubes, etc. Like this $$ S = \sum_{p=2}^\infty \sum_{n = 1}^\infty \frac{1}{n^p} $$ From the definition of the Riemann-Zeta function, we get that $$ S = \sum_{p=2}^\infty \zeta(p) $$ But I don't know if it is known if this sum converges? For the even-integer arguments, it seems like the terms are dropping off fast enough, just by a quick look at the numbers. However, I guess not much is known about the odd-integer arguments?

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Since $\zeta(p)\to1$ as $p\to\infty$ the series diverges. However, the series $$\sum_{n=2}^\infty(\zeta(n)-1)$$ does converge to $1$.

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