An Expression of $\ln(2)$ I saw online that the following infinite series has a value of $\ln(2)$:
$\sum_{n=0}^\infty \left(\dfrac{1}{n+1}-\dfrac{1}{n+2}
+\dfrac{1}{n+3}-\cdot\cdot\cdot\right)^2$
I started off by defining
$b_n=\sum_{k=0}^\infty \dfrac{(-1)^k}{n+k+1}$
I see that $b_0=ln(2)$ but that didn't help me in finding the sum of the 
original question.

Rewrite as
$$ \begin{align}S &=\sum_{n=0}^{\infty} \left [\int_0^1 dt \left (t^n - t^{n+1} + t^{n+2} - \cdots \right ) \right ]^2 \\ &=\sum_{n=0}^{\infty} \left [\int_0^1 dt \frac{t^n}{1+t} \right ]^2 \\ &= \sum_{n=0}^{\infty} \int_0^1 du \frac{u^n}{1+u} \, \int_0^1 dv \frac{v^n}{1+v} \\&= \int_0^1 \frac{du}{1+u} \, \int_0^1 \frac{dv}{1+v} \sum_{n=0}^{\infty} (u v)^n \\ &= \int_0^1 \frac{du}{1+u} \, \int_0^1 \frac{dv}{1+v} \frac1{1-u v} \\ &= \int_0^1 \frac{du}{(1+u)^2} [\log{2} - \log{(1-u)}] \\ &= \frac12 \log{2} - \int_0^1 du \frac{\log{u}}{(2-u)^2}\\ &= \frac12 \log{2} + \frac12 \log{2}\end{align}$$
Thus, $S=\log{2}$ as conjectured.
Note that
$$\begin{align}\int_0^1 \frac{dv}{1+v} \frac1{1-u v}  &= \frac1{1+u}\int_0^1 dv \left (\frac{u}{1-u v} + \frac1{1+v} \right )\\ &= -\frac{\log{(1-u)}}{1+u}+\frac{\log{2}}{1+u} \end{align}$$
and
$$\begin{align}\int_0^1 du \frac{\log{u}}{(2-u)^2} &= \frac14 \sum_{k=0}^{\infty}(k+1) \frac1{2^k} \int_0^1 du \, u^k \log{k} \\ &= -\frac14 \sum_{k=0}^{\infty}(k+1) \frac1{2^k}\frac1{(k+1)^2} \\ &= -\frac12 \sum_{k=1}^{\infty} \frac{(1/2)^k}{k} \\ &= \frac12 \log{(1/2)} \end{align}$$
