The Lerch Transcendent is defined here as

$$\Phi(z,s,a):=\sum_{k=0}^\infty \frac{z^k}{(k+a)^s}.$$

I am interested in the case $z=\frac 12,$ $s=1.$ The following limit showed up in estimating uniform distributions on an interval of length 1 under the KL divergence loss metric from samples drawn from this distribution:

$$\lim_{n\to\infty} n^3\left(\frac{\Phi(\frac 12,1,n)}{2(n-1)}-\frac{\Phi(\frac 12,1,n-1)}{2n}\right).$$

I would like to know what this limit evaluates to. I tried Mathematica which couldn't answer it. If you would like more details about the motivation, I can provide them. Thanks!