3
$\begingroup$

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!

$\endgroup$

2 Answers 2

5
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.