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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?${}$2012-11-14
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    @GerryMyerson Very fascinating! :)2012-11-14
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    In what way?${}$2012-11-14
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    @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

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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 $0$.

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    What is $B'(n)$? Just curious.2012-11-14
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    What is the meaning of the derivative of $B_n$ with respect to the integer parameter $n$?2012-11-14
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    @did change $n$ to $x$. It will be ok2012-11-14
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    @Norbert Sorry but no.2012-11-14
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    @did What is the meaning of "meaning"?2012-11-14
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    @did I plot the garph of $B(n)$ with $n\in\mathbb{R}_+$. It is ok.2012-11-14
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    @Norbert So what is $B_0'$?2012-11-15
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    $$B'_0=\lim_{n\to 0}\frac{B(n)-1}{n}$$2012-11-15
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    Now, what use is such a formula?2012-11-15
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    @Spenser What is the meaning of "play with formulas equivalent to the original question and bringing no new information nor understanding to it"?2012-11-15
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    @Norbert I do not know what you are trying to say. Sorry.2012-11-15
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    @did Complete noob here. So may I ask you to explain why the two formulas are equivalent? I have got a suspicion it is so because it is basically just replacing formulas with numbers? Am I correct? For example, is this similar to saying that $\sum_{n=0}^\infty \frac{1}{n!} = e$?2012-11-15
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    @JayeshBadwaik Because $k^x=\exp(x\log k)$ hence the derivative of the function $u_k:x\mapsto k^x$ with respect to $x$ at some $x$ is $u_k'(x)=\exp(x\log k)\log k=k^x\log k$, in particular $u_k'(0)=\log k$.2012-11-15
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    @did Sorry, you misunderstood me. I wanted to ask the reason for your comment "What is the meaning of "play with formulas equivalent to the original question and bringing no new information nor understanding to it"? "2012-11-15
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    @Spenser you do understand that you can't take $n \rightarrow 0$ because $n$ is discrete, right? And I still don't get why $B_0' = \frac {0.60378 \ldots} e$2012-11-16
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    I will answer to some of the questions that were asked to me in the following 3-posts-comment. Please remember that I do this really for the fun of argumentation. Feel free to contradict me ! :-) It would be great if the OP leave some comments also.2012-11-16
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    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.2012-11-16
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    As an analogy, recall that raising a number to an irrational exponent is really only defined by $x^r:=e^{r\log x}$. Without this natural - but nonetheless forced - extension of "exponentiation" and "root extraction", $2^{\sqrt{2}}$ would remains meaningless.2012-11-16
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    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