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This question relates to a statistics question on CrossValidated.SE where the following summation appeared in an answer:

$$I(n) \equiv \ln n + \sum_{k=1}^{n} {n \choose k} (-1)^k \ln k \quad \quad \quad \quad \text{for } n \in \mathbb{N}.$$

Does anyone know of any simpler form for this expression? (I doubt it, but thought I'd check with you brilliant people.) Are there any useful bounds/approximations/asymptotic results for this expression?

joriki
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Ben
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  • This might be helpful -- you can transform with $k=n-j$ and subtract $n\log n$, but then you have the case $n=m$ whereas the asymptotic analysis there assumes $n\gt m$. Still, you might be able to adapt the analysis. – joriki Jul 16 '18 at 05:11
  • By the way, I get that question suggested as one of the first when I enter your title in "Ask a question". You must have gotten the same display, no? – joriki Jul 16 '18 at 05:14

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The large-$n$ asymptotics of this sum are discussed in my paper Asymptotics of Certain Sums Required in Loop Regularisation. I found that the formula $$ m\int_0^{\infty} e^{-y}(1-e^{-y})^{m-1}\log{y} \, dy = -\gamma - \sum_{k \geq 1} (-1)^{k-1} \binom{n}{k} \log{k} $$ was most useful: one can apply some standard asymptotic analysis (essentially an extension of Laplace's Method) to this integral to obtain the asymptotic expansion $$ \sum_{k \geq 1} (-1)^{k} \binom{n}{k} \log{k} \sim \log{\log{n}}-\gamma - \sum_{r=1}^{\infty} \frac{1}{r}\Gamma^{(r)}(1) (\log{n})^{-r}, $$ where $\Gamma^{(r)}$ is the $r$th derivative of the $\Gamma$-function.

It seems unlikely that a closed form exists. Closed forms in terms of polynomials in harmonic numbers do exist for sums of this sort with the logarithm replaced by negative integer powers of $k$, by extending some results of Euler (which is also done in my paper), but there's no way to extend the technique to other cases, at least as far as I know: the power being a negative integer is crucial in that case.

Chappers
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