Evaluation of $\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=2-2\ln(2)$ I came across the following statements
$$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=2-2\ln 2 \qquad \tag{1}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(3 n+1)}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}} \qquad \tag{2}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(4 n+1)}=4-\frac{\pi}{2}-3 \ln 2 \qquad \tag{3}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(6 n+1)}=6-\frac{\sqrt{3} \pi}{2}-\frac{3 \ln 3}{2}-2 \ln 2 \qquad \tag{4}$$
The (1) by partial fractions
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{2}{2n+1}$$
$$=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}$$
Recall the Digamma function
$$\psi(x+1)=\gamma+\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+x}$$
Therefore
$$\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}=\psi(1+\frac{1}{2})-\gamma$$
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\psi\left(\frac{3}{2}\right)-\gamma$$
In the same token we can derive the relation for the other three ralations. My Question is: can we calculate the values of the digamma function for those values without resorting in the Gauss´s Digamma formula?
$$\psi\left(\frac{r}{m}\right)=-\gamma-\ln (2 m)-\frac{\pi}{2} \cot \left(\frac{r \pi}{m}\right)+2 \sum_{n=1}^{\left\lfloor\frac{m-1}{2}\right\rfloor} \cos \left(\frac{2 \pi n r}{m}\right) \ln \sin \left(\frac{\pi n}{m}\right)$$

I tried this approach also, but I think the resulting integral is divergent $$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1}x^{2n}dx=\int_{0}^{1}\sum_{n=1}^{\infty} \frac{x^{2n}}{n}=-\int_{0}^{1}\ln(1-x^2)dx $$

The correct answer should be $2-2\log 2$. Note that
$$
\sum\limits_{n = 1}^\infty  {\frac{1}{{n(2n + 1)}}}  = \mathop {\lim }\limits_{N \to  + \infty } \left( {\sum\limits_{n = 1}^N {\frac{1}{n}}  - 2\sum\limits_{n = 1}^N {\frac{1}{{2n + 1}}} } \right).
$$
Now, in terms of the harmonic numbers,
\begin{align*}
\sum\limits_{n = 1}^N {\frac{1}{n}}  - 2\sum\limits_{n = 1}^N {\frac{1}{{2n + 1}}} & = \sum\limits_{n = 1}^N {\frac{1}{n}}  - 2\left( {-1+\sum\limits_{n = 1}^{2N + 1} {\frac{1}{n}}  - \sum\limits_{n = 1}^N {\frac{1}{{2n}}} } \right) \\ & =2+ 2\sum\limits_{n = 1}^N {\frac{1}{n}}  - 2\sum\limits_{n = 1}^{2N + 1} {\frac{1}{n}}  =2+ 2H_N  - 2H_{2N + 1} .
\end{align*}
To finish the proof, use the asymptotics $H_k  = \log k + \gamma  + o(1)$ as $k\to +\infty$.
