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.
 A: 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
$$\begin{align}\int_0^1 du \frac{\log{u}}{(2-u)^2} &= \frac14 \sum_{k=0}^{\infty}(k+1) \frac1{2^k} \int_0^1 du \, u^k \log{k} \\ &= -\frac14 \sum_{k=0}^{\infty}(k+1) \frac1{2^k}\frac1{(k+1)^2} \\ &= -\frac12 \sum_{k=1}^{\infty} \frac{(1/2)^k}{k} \\ &= \frac12 \log{(1/2)} \end{align}$$
A: We can evaluate the sum of interest $S$, by straightforward analysis of the series.  To proceed, we write
$$\begin{align}
S&=\sum_{n=0}^\infty\left(\sum_{j=1}^\infty\frac{(-1)^{j-1}}{n+j}\,\sum_{k=1}^\infty \frac{(-1)^{k-1}}{n+k}\right) \tag 1\\\\
&=\sum_{n=0}^\infty\left(\sum_{j=n+1}^\infty\frac{(-1)^{j}}{j}\,\sum_{k=n+1}^\infty \frac{(-1)^{k}}{k}\right) \tag 2\\\\
&=\sum_{j=1}^\infty\frac{(-1)^j}{j}\left(\sum_{n=0}^{j-1}\,\sum_{k=n+1}^\infty \frac{(-1)^{k}}{k}\right)\tag 3\\\\
&=\sum_{j=1}^\infty\frac{(-1)^j}{j}\left(\sum_{n=0}^{j-1}\left(\sum_{k=n+1}^j \frac{(-1)^{k}}{k}+\sum_{k=j+1}^\infty \frac{(-1)^{k}}{k}\right)\right) \tag 4\\\\
&=\sum_{j=1}^\infty\frac{(-1)^j}{j}\left(\sum_{k=1}^{j}\frac{(-1)^k\,k}{k}+\sum_{k=j+1}^{\infty}\frac{(-1)^k\,j}{k}\right) \tag 5\\\\
&=\sum_{j=\text{odd}}^\infty\frac{(-1)^{j-1}}{j}+\sum_{j=\text{even}}^\infty\frac{(-1)^{j-1}}{j}\tag 6\\\\
&=\sum_{j=1}^\infty \frac{(-1)^{j-1}}{j} \tag 7\\\\
&=\log(2)
\end{align}$$
as was to be shown!

NOTES:
In going from $(1)$ to $(2)$, we tacitly enforced the substitutions $j\to j-n$ and $k\to k-n$.
In going from $(2)$ to $(3)$, we changed the order of summation on the $n$ and $j$ indices.
In going from $(3)$ to $(4)$, we split the summation over $k$.
In going from $(4)$ to $(5)$, we changed the order of summation on the $n$ and $k$ indices and summed the resulting summations over $n$.
In going from $(5)$ to $(6)$, we observed the sum $\sum_{k=1}^j(-1)^k$ is $-1$ for odd $j$ and $0$ otherwise.  Analogously, after changing the order of summation on the second term (and interchanging the dummy indices) on the right-hand side of $(5)$, we observed that the sum $\sum_{k=1}^{j-1}(-1)^k$ is $-1$ when $j$ is even and $0$ otherwise.
In going from $(6)$ to $(7)$ we reassemble the even and odd components of the resulting series.
A: Here's an elementary solution:
Consider expanding everything out to get a sum of elements of the form $\pm \frac1i \cdot \frac1j$. Fix $i$ and $j$ for $i\leq j$, and consider all occurrences of $\pm\frac1i \cdot \frac1j$. If $i<j$, then this occurs twice in each of the first $i$ original terms, each with sign $(-1)^{j-i}$, so they sum up to $(-1)^{j-i}\cdot\frac2j$. If $i=j$, then this occurs once in each of the first $j$ original terms, each with sign $+1$, so they sum up to $\frac1j$.
Now let's fix $j$ and aggregate all $\pm\frac1i \cdot \frac1j$ with $i\leq j$. This value is $\displaystyle\sum\limits_{i=1}^{j-1} (-1)^{j-i}\cdot\frac2j + \frac1j$, which is easily seen to be just $(-1)^{j-1}\cdot\frac1j$. Finally, we sum over all $j$ to get $\displaystyle\sum\limits_{j=1}^\infty (-1)^{j-1} \cdot \frac 1j$, which is your $b_0$.
