# Information on the sum $\sum_{n=1}^\infty \frac{\log n}{n!}$

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?

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Fascinating?${}$ –  Gerry Myerson Nov 14 '12 at 22:25
@GerryMyerson Very fascinating! :) –  Argon Nov 14 '12 at 22:26
In what way?${}$ –  Gerry Myerson Nov 14 '12 at 22:47
@GerryMyerson I don't know. Just when I see an elegant looking infinite sum, I find it very fascinating. –  Argon Nov 14 '12 at 22:48

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 $0$.
@did change $n$ to $x$. It will be ok –  Norbert Nov 14 '12 at 22:52
@did I plot the garph of $B(n)$ with $n\in\mathbb{R}_+$. It is ok. –  Norbert Nov 14 '12 at 23:25
I absolutely agree that if Bell numbers are viewed strictly as a sequence of integers - a function $B:\mathbb{N}\to\mathbb{N}$ - then my answer has no meaning at all. The derivative is just not defined, you are right. So the question is really: Is it meaningful to extend Bell numbers' definition for non-integer values? This falls into the category of opinions, but I think yes. And we have a well defined way of doing it with Dobinski's formula. –  Spenser Nov 16 '12 at 22:15
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. –  Spenser Nov 16 '12 at 22:16