Lerch transcendent: $\text{LerchPhi}^{(0,1,0)}\left(\frac{1}{2}, 0, 2\right)$ While messing around with something I got a result on WolframAlpha with a notation like this $$\text{LerchPhi}^{(0,1,0)}\left(\frac{1}{2}, 0, 2\right)$$
I know that $$\text{LerchPhi}(z,s,a)=\sum_{k=0}^\infty\frac{z^k}{(a+k)^s}$$
But I don't understand the above notation,and I'm not able to reproduce it on wolfram alpha.
 A: In this particular case, using 
$$
\text{LerchPhi}(z,s,a)=\sum_{k=0}^\infty\frac{z^k}{(a+k)^s},
$$
you may write 
$$
\begin{align}
\text{LerchPhi}^{(0,1,0)}\left(\frac{1}{2}, 0, 2\right) & = \frac{\partial}{\partial s}
\left.\left( \sum_{k=0}^\infty\frac{1}{ 2^k(k+2)^s} \right)  \right|_{s=0} 
\\\\ & =\sum_{k=0}^\infty\frac{1}{ 2^k}\frac{\partial}{\partial s}
\left.\left( \frac{1}{(k+2)^s} \right) \right|_{s=0}
\\\\ & =\sum_{k=0}^\infty \frac{1}{ 2^k}
\left. \left(- \frac{\log(k+2)}{ (k+2)^s} \right) \right|_{s=0}
\\\\ & =-\sum_{k=0}^\infty
\frac{\log(k+2)}{ 2^k} 
\\\\ & =-4\sum_{k=1}^\infty
\frac{\log(k)}{ 2^k} 
\\\\ & =-4 \log \left(\prod_{k=1}^\infty
k^{1/2^k} \right)
\\\\ & =-4 \log \sigma
\end{align}
$$
where 
$$
\displaystyle \sigma = \prod_{k=1}^\infty
k^{1/2^k} = \sqrt{1\sqrt{2\sqrt{3\cdots}}}
$$ is known as the Somos quadratic recurrence constant.
Observe that, upon writing $\displaystyle \sigma = \frac{\sigma^2}{\sigma}$, you also have
$$
\displaystyle \sigma =  \left(\frac{2}{1}\right)^{1/2}\left(\frac{3}{2}\right)^{1/4}\left(\frac{4}{3}\right)^{1/8} \cdots
$$
which converges faster than the first product.
