Harmony in the world!

There is a typo in the question: the sum should be from $1$ to $\infty$ , and not from $0$ to $\infty$ .

Starting from the generating function for the harmonic numbers $\sum_{n-1}^\infty H_n x^n \; = \; -\frac{\ln(1-x)}{1-x} \qquad \qquad |x| < 1 \;,$ we obtain $\sum_{n=1}^\infty \frac{H_n}{n} x^n \; = \; -\int_0^x \frac{\ln(1-u)}{u(1-u)}\,du \qquad \qquad |x| < 1 \;,$ and hence that $\sum_{n=1}^\infty \frac{(-1)^{n-1}H_n}{n}x^n \; = \; \int_0^x \frac{\ln (1+u)}{u(1+u)}\,du \qquad \qquad |x| < 1 \;.$ Letting $x \to 1$ we deduce that $\begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}H_n}{n} & = & \displaystyle \int_0^1 \frac{\ln(1+u)}{u(1+u)}\,du \\ & = & \displaystyle \int_0^1 \ln(1+u)\left[ \frac{1}{u} - \frac{1}{1+u}\right]\,du \\ & = & \displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} \int_0^1 u^{n-1}\,du - \int_0^1 \frac{\ln(1+u)}{1+u}\,du \\ & = & \displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} - \Big[\tfrac12\ln(1+u)^2\Big]_0^1 \\ & = & \tfrac12\zeta(2) - \tfrac12(\ln2)^2 \end{array}$ so that $P=R=2$ , making the answer $\boxed{4}$ .