\displaystyle\sum_{k=1}^{\infty}\frac{q^k}{k(k+1)}=\sum_{k=1}^{\infty}(\frac{q^k}{k}-\frac{q^k}{k+1})=\sum_{k=1}^{\infty}\frac{q^k}{k}-\frac{1}{q}(\sum_{k=1}^{\infty}\frac{q^{k+1}}{k+1}+q-q)=\sum_{k=1}^{\infty}\frac{q^k}{k}-\frac{1}{q}(\sum_{k=1}^{\infty}\frac{q^{k+1}}{k+1}+q)+1=\sum_{k=1}^{\infty}\frac{q^k}{k}-\frac{1}{q}\sum_{m=1}^{\infty}\frac{q^m}{m}+1=\ln\frac{1}{1-q}-\frac{1}{q}\ln\frac{1}{1-q}+1=1-\frac{q}{q}\ln(1-q)+\frac{1}{q}\ln(1-q)=1+\frac{(1-q)\ln(1-q)}{q}