Value of $\sum_\limits{n=1}^{\infty}\frac1{n(2n+1)}$ I must find the sum $$\sum_\limits{n=1}^{\infty}\frac1{n(2n+1)}$$ I have already tried partial fractions, but obtained an indeterminate form of infinity minus infinity. Perhaps completing the square would help?
 A: Note that
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \log{2} $$
Then the desired sum is equal to
$$2\sum_{n=1}^{\infty} \left (\frac1{2 n}-\frac1{2 n+1} \right ) = 2 \left (\frac12 - \frac13 + \frac14 - \frac15 +\cdots \right ) = 2 (1-\log{2})$$
A: $$\sum_{n=1}^{+\infty}\frac{1}{n(2n+1)}=2\sum_{n=1}^{+\infty}\int_{0}^{1}\left(x^{2n-1}-x^{2n}\right)\,dx = 2\int_{0}^{1}\frac{x}{1+x}\,dx = 2-\log 4.$$
A: For $|x|<1$, it holds $$\sum_{k=1}^\infty \frac{x^k}{k}=-\ln(1-x)$$
Hence, for $|x|<1$, it holds $$\sum_{k=1}^\infty \frac{x^{2k}}{k}=-\ln(1-x^2)=-\ln(1-x)-\ln(1+x)$$
Hence, for $|x|<1$ it holds $$\sum_{k=1}^\infty \frac{x^{2k+1}}{k(2k+1)}=-\int_0^x\ln(1-t^2)\,dt=(1-x)\ln(1-x)-(1+x)\ln(1+x)+2x$$
Since the sum converges for $x=1$ and $$\lim_{x\to1^-}\left[(1-x)\ln(1-x)-(1+x)\ln(1+x)+2x\right]=2-2\ln2$$ it holds $$\sum_{k=1}^\infty \frac{1}{k(2k+1)}=2-2\ln2$$
A: We have $$\sum_{n=1}^{\infty} \frac{1}{n(2n+1)}=\sum_{n=0}^{\infty} \frac{1}{2(n+1)(n+\frac{3}{2})} = \sum_{n=0}^{\infty} \frac{1}{n+1}-\frac{1}{n+\frac{3}{2}}=\psi(\frac{3}{2})+\gamma = 2+\psi(\frac{1}{2})+\gamma = 2-\ln 4$$
