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.

Rewrite the series as follows:

$$\sum_{n=1}^\infty\frac{1}{4n^2+2n}=\sum_{n=1}^\infty\frac{1}{2n(2n+1)}=\sum_{n=1}^\infty\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=\sum_{m=2}^\infty\frac{(-1)^m}{m}.$$

You may now evaluate it using the Taylor series of logarithm.

Added: The third equality is justified as follows:

Write $a_N=\sum_{n=1}^N\left(\dfrac{1}{2n}-\dfrac{1}{2n+1}\right)$ and $b_M=\sum_{m=2}^M\dfrac{(-1)^m}{m}$. The sequence of partial sums $b_M$ is convergent by the Leibniz criterion. Furthermore, $a_N = b_{2N+1}$ holds for all $N\in\mathbb N$, i.e. $(a_N)_{N=1}^\infty$ is a subsequence of $(b_M)_{M=1}^\infty$. Therefore these sequences converge to the same limit, which justifies the equality.

$$\dfrac1{4n^2+2n} = \dfrac1{2n(2n+1)} = \dfrac1{2n} - \dfrac1{2n+1}$$ Hence, \begin{align} \lim_{m \to \infty} \sum_{n=1}^{m} \dfrac1{4n^2+2n} & = \lim_{m \to \infty} \sum_{n=1}^{m} \left(\dfrac1{2n} - \dfrac1{2n+1} \right)\\ & = \lim_{m \to \infty}\left(\dfrac12 - \dfrac13 + \dfrac14 - \dfrac15 \pm \cdots +\dfrac1{2m} - \dfrac1{2m+1}\right)\\ & = 1 - \lim_{m \to \infty}\left(1-\dfrac12 + \dfrac13 - \dfrac14 + \dfrac15 \mp \cdots -\dfrac1{2m} + \dfrac1{2m+1}\right)\\ & = 1 - \lim_{m \to \infty}\left(1-\dfrac12 + \dfrac13 - \dfrac14 + \dfrac15 \mp \cdots -\dfrac1{2m} \right) - \lim_{m \to \infty}\dfrac1{2m+1}\\ & = 1 - \lim_{m \to \infty}\dfrac1{2m+1} - \lim_{m \to \infty} \sum_{n=1}^{2m} \dfrac{(-1)^{n-1}}{n}\\ & = 1 - 0 - \log(1+1)\,\,\, \left\{\text{Since $\log(1+x) = -\displaystyle \sum_{n=1}^{\infty}\dfrac{(-x)^{n}}n$} \right\}\\ & = 1 - \log(2) \end{align}

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

Knowing the answer is helpful. Recall that $$\ln 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots.$$ Thus $$\ln 2=1-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\cdots.$$ The "typical" term $\dfrac{1}{2k}-\dfrac{1}{2k+1}$ simplifies to $\dfrac{1}{2k(2k+1)}$.