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.
 A: $$\dfrac1{4n^2+2n} = \dfrac1{2n(2n+1)} = \dfrac1{2n} - \dfrac1{2n+1}$$
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}
A: Hint:  $\frac 1{4n^2+2n} = \frac 1{2n}-\frac 1{2n+1}$
A: 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 $\dfrac{1}{2k(2k+1)}$. 
A: 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.
A: $\gamma\,$: Euler-Mascheroni constant. $\quad\Psi\left(z\right)\,$: Digamma function.
\begin{align}
&\\[3mm]
\sum_{n = 1}^{\infty}{1 \over 4n^{2} + 2n}
&=
{1 \over 4}\sum_{n = 1}^{\infty}{1 \over n\left(n + 1/2\right)}
=
{1 \over 4}\sum_{n = 0}^{\infty}{1 \over \left(n + 1\right)\left(n + 3/2\right)}
=
{1 \over 4}{\Psi\left(3/2\right) - \Psi\left(1\right) \over 3/2 - 1}
\\[3mm]&=
{1 \over 2}
\left\lbrack\Psi\left(3 \over 2\right) - \Psi\left(1\right)\right\rbrack
\\[5mm]&
\end{align}
$$
\Psi\left(3 \over 2\right)
=
\underbrace{\Psi\left(1 \over 2\right)}
_{-\gamma\ -\ 2\ln\left(2\right)} + {1 \over \left(1/2\right)}
=
-\gamma + 2\left\lbrack 1 - \ln\left(2\right)\right\rbrack\,,
\qquad\qquad
\Psi\left(1\right) = -\gamma
$$
$$
\begin{array}{|c|}\hline\\
{\large\quad\sum_{n = 1}^{\infty}{1 \over 4n^{2} + 2n}
=
1 - \ln\left(2\right)\quad}
\\ \\
\hline
\end{array}
$$
A: Hint:
Another approach is to Complete the square in the denominator and use the Fourier transform and Poison summation formula approach.
