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?

$$\psi\left(\frac{r}{m}\right)=-\gamma-\ln (2 m)-\frac{\pi}{2} \cot \left(\frac{r \pi}{m}\right)+2 \sum\_{n=1}^{\left\lfloor\frac{m-1}{2}\right\rfloor} \cos \left(\frac{2 \pi n r}{m}\right) \ln \sin \left(\frac{\pi n}{m}\right)$$

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