A rather peculiar sum

We use that $\displaystyle \zeta(k) = \sum_{n=1}^{\infty} \frac{1}{n^k}$ and thus rearrange the series given. We start by swapping summations:

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

We then invoke the expansion $\displaystyle -\ln(1-x) = \sum_{k=1}^{\infty} \frac{x^k}{k}$ .

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

Following that, we proceed to expose the implicit limit:

$\sum_{n=2}^{\infty} -\frac{1}{n}-\ln(1-\frac{1}{n}) = \lim_{N \to \infty} \sum_{n=2}^{N} -\frac{1}{n}-\ln(1-\frac{1}{n}) = \lim_{N \to \infty} 1-H_N-\sum_{n=2}^{N} \ln(1-\frac{1}{n})$

where $\displaystyle H_N = \sum_{n=1}^{N} \frac{1}{n}$ is the Nth harmonic number. The obtained sum, $\displaystyle \sum_{n=2}^{N} \ln(1-\frac{1}{n})$ , is a telescoping series:

$\sum_{n=2}^{N} \ln(1-\frac{1}{n}) = \ln \left[ \prod_{n=2}^{N} \left( 1-\frac{1}{n} \right) \right] = \ln(\frac{1}{2} \frac{2}{3} \ldots \frac{N-1}{N}) = \ln(\frac{1}{N}) = -\ln N$

Thus, our limit is

$\lim_{N \to \infty} 1-H_N-\sum_{n=2}^{N} \ln(1-\frac{1}{n}) = \lim_{N \to \infty} 1-H_N+\ln N = 1-\gamma$

where $\gamma$ is the Euler-Mascheroni constant, defined by $\displaystyle \gamma = \lim_{N \to \infty} H_N-\ln N = 0.5772\ldots$ . Hence,

$\sum_{k=2}^{\infty} \frac{\zeta(k)-1}{k} = 1-\gamma \approx \boxed{0.423}$