Closed form representation of alternating series 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}

\begin{align}
\sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k+1}} + \sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k}}
 &=
\sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k+1}} + \sum_{k=0}^\infty{\frac{(-1)^{k}}{k+1}}
\\ &=
\sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k+1}} + \sum_{k=1}^\infty{\frac{(-1)^{k}}{k+1}}+1
\\ &=1
\end{align}
