Find the value of $\sum_{n=1}^{\infty} \frac{2}{n}-\frac{4}{2n+1}$ Find the value of $$S=\sum_{n=1}^{\infty}\left(\frac{2}{n}-\frac{4}{2n+1}\right)$$  
My Try:we have
$$S=2\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{2}{2n+1}\right)$$  
$$S=2\left(1-\frac{2}{3}+\frac{1}{2}-\frac{2}{5}+\frac{1}{3}-\frac{2}{7}+\cdots\right)$$ so
$$S=2\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\cdots\right)$$  But we know
$$\ln2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$  So
$$S=2(2-\ln 2)$$
Is this correct?

We can use the digamma function $\psi$. It is known that
$$\psi(z+1)=-\gamma+\sum_{n=1}^{\infty}\dfrac{z}{n(n+z)},\ \ \forall z \in \mathbb{C} \setminus \{-1,-2,...\}.$$
Thus for $z=\frac{1}{2}$ we obtain
$$\psi\left(\dfrac{3}{2}\right)=-\gamma+\sum_{n=1}^{\infty}\dfrac{1}{n(2n+1)}=-\gamma+\sum_{n=1}^{\infty}\left(\dfrac{1}{n}-\dfrac{2}{(2n+1)}\right)=-\gamma+\dfrac{S}{2}.$$
From the other hand, for the half-integer values of $\psi$ we have 
$$\psi\left(n+\dfrac{1}{2}\right)=-\gamma-2\ln 2+\sum_{k=1}^n\dfrac{2}{2k-1},$$
from where, for $n=1$, we conclude that
$$\psi\left(\dfrac{3}{2}\right)=-\gamma-2\ln 2+2.$$
Finally $-\gamma+\dfrac{S}{2}=-\gamma-2\ln 2+2 \Rightarrow S=4(1-\ln 2),$
as Simply Beautiful Art mentioned.
Note: S is a real number and it can be justified via comparison with the series of the Basel Problem.
