Lerch transcendent: $\text{LerchPhi}^{(0,1,0)}\left(\frac{1}{2}, 0, 2\right)$

Question

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.

Answer 1

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.