Evaluating $\sum_{n=1}^\infty\frac{1}{4n(2n+1)}$

Question

How to evaluate this sum, derived from "Lockdown math" by 3Blue1Brown?

$$\sum_{n=1}^\infty\frac{1}{4n(2n+1)}$$

Answer 1

$$\frac{1}{4n(2n+1)}=\frac{1}{4n}-\frac{1/2}{2n+1}=\frac{1}{2}\left(\frac{1}{2n}-\frac{1}{2n+1}\right)$$

Hence

$$\sum_{n=1}^{\infty} \frac{1}{ 4n(2n+1)} =\frac 12 \sum_{n=1}^{\infty} \left(\frac{1}{2n}-\frac{1}{2n+1}\right)= \frac 12\left(1-\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\right)=\ldots $$

where $${\displaystyle \ln(1+x)=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}x^{n},\quad \ln(2)=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}}$$

Answer 2

$$\frac{1}{4n(2n+1)}=\frac{1}{4n}-\frac{1}{2(2n+1)}=\frac{1}{4n}-\frac{1}{4n+2}$$ Can you see where this is going?