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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?

Argon
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1 Answers1

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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 $0$.

Spenser
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  • What is $B'(n)$? Just curious. – glebovg Nov 14 '12 at 22:47
  • What is the meaning of the derivative of $B_n$ with respect to the integer parameter $n$? – Did Nov 14 '12 at 22:50
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    @did change $n$ to $x$. It will be ok – Norbert Nov 14 '12 at 22:52
  • @Norbert Sorry but no. – Did Nov 14 '12 at 22:53
  • @did What is the meaning of "meaning"? – Spenser Nov 14 '12 at 23:08
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    @did I plot the garph of $B(n)$ with $n\in\mathbb{R}_+$. It is ok. – Norbert Nov 14 '12 at 23:25
  • @Norbert So what is $B_0'$? – Kaster Nov 15 '12 at 00:15
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    $$B'0=\lim{n\to 0}\frac{B(n)-1}{n}$$ – Spenser Nov 15 '12 at 00:27
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    Now, what use is such a formula? – fedja Nov 15 '12 at 01:14
  • @Spenser What is the meaning of "play with formulas equivalent to the original question and bringing no new information nor understanding to it"? – Did Nov 15 '12 at 07:00
  • @Norbert I do not know what you are trying to say. Sorry. – Did Nov 15 '12 at 07:01
  • @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 $\sum_{n=0}^\infty \frac{1}{n!} = e$? –  Nov 15 '12 at 09:52
  • @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 $u_k'(0)=\log k$. – Did Nov 15 '12 at 13:15
  • @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"? " –  Nov 15 '12 at 14:04
  • @Spenser you do understand that you can't take $n \rightarrow 0$ because $n$ is discrete, right? And I still don't get why $B_0' = \frac {0.60378 \ldots} e$ – Kaster Nov 16 '12 at 20:15
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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. – Spenser Nov 16 '12 at 22:12
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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. – Spenser Nov 16 '12 at 22:15
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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. – Spenser Nov 16 '12 at 22:16
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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. – Spenser Nov 16 '12 at 22:16