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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!

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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 $$ \dfrac{1}{n} - \dfrac{4}{n^2} + O(1/n^3)$$

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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 $n^{-1} - 4 n^{-2} + O(n^{-3})$.

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