I consider the following series
\begin{align}
\sum\_{n=0}^\infty{\frac{(-1)^{n+2}}{n+2}} \stackrel{?}{=} 1-\ln(2)
\end{align}
Wolfram tells me that it is equal to $1-\ln(2)$.
I know the following
\begin{align}
&\sum\_{k=1}^\infty{\frac{(-1)^{k+1}}{k}} = \ln(2)\\
&\sum\_{n=0}^\infty{\frac{(-1)^{n+2}}{n+2}} = \sum\_{k=1}^\infty{\frac{(-1)^{k+1}}{k+1}}
\end{align}
I thus need to show that the following is true, but I would not know how to do it.
\begin{align}
\sum\_{k=1}^\infty{\frac{(-1)^{k+1}}{k+1}} + \sum\_{k=1}^\infty{\frac{(-1)^{k+1}}{k}} = 1
\end{align}