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I want to find an simple analytic expression for $$\sum_{k=0}^\infty \frac{\log(k+1)}{k!}.$$ Wolfram alpha just tells me this is approximately 1.55 (for natural log).

Equivalently, this is equal to $$e \cdot \underset{K \leftarrow \mathsf{Poisson}(1)}{\mathbb{E}} \left[K \log K\right]$$ (taking the convention that $0 \log 0 = 0$ for continuity). Surely this sort of expectation has been evaluated?

This has come up in trying to analyze the entropy of a random distribution. Any help would be much appreciated.

Another answer suggests using the fact that $\mathbb{E}[K \log K] = \frac{\mathrm{d}}{\mathrm{d}x} \mathbb{E}[K^x]$ evaluated at $x=1$. However, I don't know if this brings us closer to a simple expression -- I don't have a good formula for $\mathbb{E}[K^x]$ unless $x$ is an integer.

Thomas
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  • This seems to do the job: https://math.stackexchange.com/questions/237519/information-on-the-sum-sum-n-1-infty-frac-log-nn?rq=1 – Tai Mar 15 '18 at 00:39
  • @TaisukeYasuda This is particularly interesting because the question you cite involves the Bell numbers. $B_n$ is the $n$th moment of a Poisson distribution with mean $1$, and the OP seems to be interested in this distribution. – saulspatz Mar 15 '18 at 01:03

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Due to Frullani's integral $\log(k+1)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-(k+1)x}}{x}\,dx $ and the change of variable $x=-\log t$

$$\sum_{k\geq 0}\frac{\log(k+1)}{k!} = \int_{0}^{1}\frac{e^t-e}{\log t}\,dt $$ where it is simple to provide accurate numerical approximations for the RHS, since $\frac{e^{t}-e}{\log t}$ is approximately linear on $(\varepsilon,1-\varepsilon)$ for $\varepsilon>0$ not too close to zero:

$\hspace1in$enter image description here

A better approximation of the integrand function is $e\sqrt{\frac{t(t+1)}{2}}$, for instance.

Jack D'Aurizio
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