# Sum of the Riemann-Zeta function at the integers

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?

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