A calculation involving Lerch Transcendents 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!
 A: Using $$ \dfrac{1}{n} - \dfrac{k}{n^2} + \dfrac{k^2}{n^3} + \ldots - \dfrac{k^{2j}}{n^{2j+1}}\le \dfrac{1}{n+k} \le \dfrac{1}{n} - \dfrac{k}{n^2} +
\ldots + \dfrac{k^{2j+1}}{n^{2j+2}}$$
we get
$$ \Phi(1/2,1,n) =  \dfrac{2}{n} - \dfrac{2}{n^2} + \dfrac{6}{n^3} + O(n^{-4})$$
which is enough to show that your expression is
$$ \dfrac{1}{n} - \dfrac{4}{n^2} + O(1/n^3)$$
A: Numerical computation suggests that the limit is zero and the 
expression has an asymptotic series starting $n^{-1} - 4 n^{-2} + O(n^{-3})$.
To prove this, start from
$$
\Phi(\frac12,1,n) = \sum_{k=0}^\infty \frac1{2^k(k+n)},
$$
and expand $1/(k+n)$ in a geometric series 
$1/n - k/n^2 + k^2/n^3 - k^3/n^4 + - \cdots$ to get
$$
\Phi(\frac12,1,n) 
\sim \sum_{i=0}^\infty (-1)^i \left(
   \frac1{n^{i+1}} \sum_{k=0}^\infty \frac{k^i}{2^k}
  \right)
= \frac2n - \frac2{n^2} + \frac6{n^3} - \frac{26}{n^4} + \frac{150}{n^5}
- + \cdots
$$
(the numerators are OEIS sequence
A076726; the geometric series diverges for $k \geq n$,
but the contribution of those terms to the sum decays exponentially).
Thus
$$
\Phi(\frac12,1,n)
\sim \frac2n + \frac4{n^3} - \frac{12}{n^4} + \frac{76}{n^5} \cdots
$$
and thus
$$
\frac{\Phi(\frac 12,1,n)}{2(n-1)}-\frac{\Phi(\frac 12,1,n-1)}{2n}
\sim \frac1{n^4} 
 - \frac4{n^5}  
 + \frac{27}{n^6}
 - \frac{206}{n^7}
 + \frac{1865}{n^8}
 - \frac{19440}{n^9}
\cdots .
$$
Multiplying by $n^3$ yields the observed $n^{-1} - 4 n^{-2} + O(n^{-3})$.
