Mangoldt

We have the following Laurent expansions about $s=1$ : $\begin{array}{rcl} -\frac{\zeta'(s)}{\zeta(s)} \; = \; \displaystyle\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s} & = & (s-1)^{-1} - \gamma + (\gamma^2 + 2\gamma_1)(s-1) + O\big((s-1)^2\big) \\ \zeta(s) \; = \; \displaystyle \sum_{n=1}^\infty \frac{1}{n^s} & = & (s-1)^{-1} + \gamma - \gamma_1(s-1) + O\big((s-1)^2\big) \end{array}$ where $\gamma_1$ is the first Stieltjes constant, and hence $\sum_{n=1}^\infty \frac{\Lambda(n)-1}{n^s} \; = \; -\frac{\zeta'(s)}{\zeta(s)} - \zeta(s) \; = \; -2\gamma + (\gamma^2 + 3\gamma_1)(s-1) + O\big((s-1)^2\big)$ Thus we have a removable singularity at $s=1$ for this combined function, and hence $\sum_{n=1}^\infty \frac{\Lambda(n)-1}{n} \; = \; \lim_{s\to1}\left[ -\frac{\zeta'(s)}{\zeta(s)} - \zeta(s)\right] \; = \; -2\gamma$ making the desired answer equal to $\boxed{-2}$ .