Sum of $\frac{1}{(2n)(2n+1)}$ from $1$ to infinity

Can anyone tell me about the sum of the series $$\sum_{n=1}^\infty \frac{1}{(2n)(2n+1)}?$$ This is not a usual telescoping sum in which all the terms cancel out.

Hints:

First, it really is true that

$$\frac1{(2n)(2n+1)}=\frac1{2n}-\frac1{2n+1}=\frac12-\frac13+\frac14-\frac15+\frac16-\frac17+\ldots$$

Now, check that

$$f(x)=\frac1{1+x}=1-x+x^2-x^3+\ldots\implies$$

$$\implies\log2=\int\limits_0^1\frac{dx}{1+x}=1-\frac12+\frac13-\frac14+\ldots$$

As $\frac1{(2n)(2n+1)}=\frac1{2n}-\frac1{2n+1}$ the series turns into $$ \sum_{n=2}^\infty\frac{(-1)^n}{n}.$$ You may recognize the Taylor series $$\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}$$ that is valid also for $x=-1$, so the result is $1-\ln 2$.

$$\frac{1}{(2n)(2n+1)}=\frac{1}{2n}-\frac{1}{2n+1}$$