Evaluating $\sum_{n=1}^\infty \frac{7}{n(2n-1)}-\frac{3}{7^{n+1}}$

Question

Find limit of this series: $$\sum_{n=1}^\infty \frac{7}{n(2n-1)}-\frac{3}{7^{n+1}}$$

So far I got to

$$\sum_{n=1}^\infty \frac{7}{n(2n-1)}- \sum_{n=1}^\infty \frac{3}{7^{n+1}}$$ where $$\sum_{n=1}^\infty \frac{3}{7^{n+1}} = \frac{7}{12}$$

Answer 1

Hint:

$$\dfrac2{2n(2n-1)}=\dfrac1{2n-1}-\dfrac1{2n}=-\dfrac{x^{2n-1}}{2n-1}-\dfrac{x^{2n}}{2n}$$ where $x=-1$

Now for $-1\le x<1,$ $$\ln(1-x)=-\sum_{r=1}^\infty\dfrac{x^r}r$$

See What is the correct radius of convergence for $\ln(1+x)$?