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Explore Stack InternalActually it suffices to know the generating function
$$\sum_{k\geq 1}H^{(p)}_kx^k=\frac{\mathrm{Li}_p(x)}{1-x}$$
Upon integrating we obtain
$$\sum_{k\geq 1}\frac{H^{(p)}_k}{k}x^k=\mathrm{Li}_{p+1}(x)+\int^x_0 \frac{\mathrm{Li}_p(t)}{1-t}\,d t$$
$$\sum_{k\geq 1}\frac{H_k}{k}x^k=\mathrm{Li}_{2}(x)+\frac{1}{2}\log^2(1-x)$$
$$\sum_{k\geq 1}\frac{H_k}{k}(-x)^k=-\frac{\pi^2}{12}+\frac{1}{2}\log^2(2)$$$$\sum_{k\geq 1}\frac{H_k}{k}(-1)^k=-\frac{\pi^2}{12}+\frac{1}{2}\log^2(2)$$
Actually it suffices to know the generating function
$$\sum_{k\geq 1}H^{(p)}_kx^k=\frac{\mathrm{Li}_p(x)}{1-x}$$
Upon integrating we obtain
$$\sum_{k\geq 1}\frac{H^{(p)}_k}{k}x^k=\mathrm{Li}_{p+1}(x)+\int^x_0 \frac{\mathrm{Li}_p(t)}{1-t}\,d t$$
$$\sum_{k\geq 1}\frac{H_k}{k}x^k=\mathrm{Li}_{2}(x)+\frac{1}{2}\log^2(1-x)$$
$$\sum_{k\geq 1}\frac{H_k}{k}(-x)^k=-\frac{\pi^2}{12}+\frac{1}{2}\log^2(2)$$
Actually it suffices to know the generating function
$$\sum_{k\geq 1}H^{(p)}_kx^k=\frac{\mathrm{Li}_p(x)}{1-x}$$
Upon integrating we obtain
$$\sum_{k\geq 1}\frac{H^{(p)}_k}{k}x^k=\mathrm{Li}_{p+1}(x)+\int^x_0 \frac{\mathrm{Li}_p(t)}{1-t}\,d t$$
$$\sum_{k\geq 1}\frac{H_k}{k}x^k=\mathrm{Li}_{2}(x)+\frac{1}{2}\log^2(1-x)$$
$$\sum_{k\geq 1}\frac{H_k}{k}(-1)^k=-\frac{\pi^2}{12}+\frac{1}{2}\log^2(2)$$
Actually it suffices to know the generating function
$$\sum_{k\geq 1}H^{(p)}_kx^k=\frac{\mathrm{Li}_p(x)}{1-x}$$
Upon integrating we obtain
$$\sum_{k\geq 1}\frac{H^{(p)}_k}{k}x^k=\mathrm{Li}_{p+1}(x)+\int^x_0 \frac{\mathrm{Li}_p(t)}{1-t}\,d t$$
$$\sum_{k\geq 1}\frac{H_k}{k}x^k=\mathrm{Li}_{2}(x)+\frac{1}{2}\log^2(1-x)$$
$$\sum_{k\geq 1}\frac{H_k}{k}(-x)^k=-\frac{\pi^2}{12}+\frac{1}{2}\log^2(2)$$
