View source for Z2F1(1,1;2,-z) equals log(1+z)

==Theorem==
The following formula holds:
$$\log(1+z)=z{}_2F_1(1,1;2;-z),$$
where $\log$ denotes the [[logarithm]] and ${}_2F_1$ denotes the [[hypergeometric pFq]].

==Proof==
Calculate 
$$\begin{array}{ll}
z{}_2F_1(1,1;2;-z) &= z\displaystyle\sum_{k=0}^{\infty} \dfrac{1^{\overline{k}}1^{\overline{k}}}{2^{\overline{k}}k!} (-z)^k \\
&= \displaystyle\sum_{k=0}^{\infty} \dfrac{\left( \frac{\Gamma(k+1)}{\Gamma(1)} \right)^2}{\left( \frac{\Gamma(2+k)}{\Gamma(2)} \right)k!}(-1)^k z^{k+1} \\
&= \displaystyle\sum_{k=0}^{\infty} \dfrac{(k!)^2(-1)^k}{(k+1)!k!} z^{k+1} \\
&= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k+1} z^{k+1} \\
&= -\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k z^k}{k} \\
&= \log(1+z),
\end{array}$$
using a well-known formula for the [[logarithm|Taylor series of $\log(1+z)$]]. █ 

==References==