9
$\begingroup$

In my personal study of interesting sums, I came up with the following sum that I could not evaluate:

$\sum_{n=1}^\infty \frac{\log n}{n!} = 0.60378\dots$

I would be very interested to see what can be done to this sum. Does a closed form of this fascinating sum exist?

  • 6
    @GerryMyerson I don't know. Just when I see an elegant looking infinite sum, I find it very fascinating.2012-11-14

1 Answers 1

5

Using Dobinski's formula for Bell numbers, we have $B(n)=\frac{1}{e}\sum_{k=0}^{\infty}\frac{k^n}{k!}$ Hence, $\frac{d}{dn}B(n)=\frac{1}{e}\sum_{k=2}^{\infty}\frac{k^n\log k}{k!}$ whence, $\sum_{k=1}^{\infty}\frac{\log k}{k!}=B'_0 e$ Note that the first term ($k=1$) is $0$.

  • 2
    With that said, isn't it just pretty to notice that this "fascinating sum" is precisely $e$ times the slope at the origin of the natural extension of Bell numbers? I think it is, that's all. Why? Because it links two apparently unrelated concepts together. Is it "useful"? "meaningful"? For the development of mathematics, I don't know. But at least, it is useful for the people who look at it and find it somewhat interesting or just pretty.2012-11-16