Evaluating the series $\sum\limits_{n=1}^\infty \frac1{4n^2+2n}$ How do we evaluate the following series:
$$\sum_{n=1}^\infty \frac1{4n^2+2n}$$
I know that it converges by the comparison test. Wolfram Alpha gives the answer $1 - \ln(2)$, but I cannot see how to get it. The Taylor series of logarithm is nowhere near this series.

Hint:  $\frac 1{4n^2+2n} = \frac 1{2n}-\frac 1{2n+1}$
