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 $$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1}x^{2n}dx=\int_{0}^{1}\sum_{n=1}^{\infty} \frac{x^{2n}}{n}=-\int_{0}^{1}\ln(1-x^2)dx $$

$$
\begin{align}
\sum_{n=1}^\infty\frac1{n(2n+1)}
&=2\lim_{N\to\infty}\sum_{n=1}^N\left(\frac1{2n}-\frac1{2n+1}\right)\tag1\\
&=2\lim_{N\to\infty}\sum_{n=1}^N\left(\frac1n-\left(\frac1{2n}+\frac1{2n+1}\right)\right)\tag2\\
&=2\lim_{N\to\infty}\left(\sum_{n=1}^N\frac1n-\sum_{n=2}^{2N+1}\frac1n\right)\tag3\\
&=2\lim_{N\to\infty}\left(1-\sum_{n=N+1}^{2N+1}\frac1n\right)\tag4\\
&=2\lim_{N\to\infty}\left(1-\sum_{n=N+1}^{2N+1}\frac Nn\frac1N\right)\tag5\\
&=2\left(1-\int_1^2\frac{\mathrm{d}x}x\right)\tag6\\[9pt]
&=2(1-\log(2))\tag7
\end{align}
$$
Explanation:
$(1)$: partial fractions
$\phantom{\text{(1):}}$ write infinite sum as a limit of partial sums
$(2)$: arithmetic
$(3)$: combine the inner parenthetical terms into one sum
$(4)$: cancel terms
$(5)$: multiply by $\frac NN$
$(6)$: convert Riemann Sum into integral
$(7)$: evaluate integral
