I checked that calculation with Mathematica, and it is indeed correct. Why does it converge?
It's essentially
\( \displaystyle{ sld(n) = \prod_{k=0}^\infty (n+k)^{1/2^{k+1}} } \)
Turning this product into a sum with logarithms gives
\( \displaystyle{ \log sld(n) = \sum_{k=0}^\infty \frac{1}{2^{k+1}} \log(n+k) } \)
That infinite series can easily be shown to converge.
After a lot of searching, I found
Lerch zeta function with the "Lerch transcendent"
\( \displaystyle{ \Phi(z,s,a) = \sum_{k=0}^\infty \frac{z^k}{(k+a)^s} } \)
Take the derivative with respect to s:
\( \displaystyle{ \frac{d}{ds} \Phi(z,s,a) = - \sum_{k=0}^\infty \frac{z^k}{(k+a)^s} \log(k+a) } \)
So
\( \displaystyle{ \log sld(n) = - \frac12 \left. \frac{d}{ds} \Phi(1/2,s,n) \right|_{s=0} } \)
Mathematica has the LerchPhi function, and was able to calculate its derivative.
