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As in the title, I'm in quest for $\sum_{n=1}^\infty \log(n)\cdot x^n$, where $0 \le x \lt 1$

Wolfram Alpha says: $-\operatorname{PolyLog}^{(1, 0)}(0, x)$, but I don't understand what that means. (Of course, PolyLog stays for the polylogarithm).

Background

It's about "how many bits I need to encode a real number $0 < r < 1$ with a tolerance $\delta/2$? The "naive" response is $-log_2(\delta)$.

Nevertheless (long story short) I need a different approach:

  1. I can encode every positive integer $n$ with approximately $C\cdot\log(n)$ bits

  2. Let $0 < x_i < 1$ be a pseudo-random sequence, and let $N$ be the 1st index so that $r-\delta/2 <x_N< r+\delta/2$. Then let's say that we can transmit $r$ via $N$ (with the tolerance $\delta$). So we need $C\cdot\log(N)$ bits...

  3. But then I need the expected value $E(C\cdot\log(N)) = \sum_{n=1}^\infty C\cdot\log(n)\cdot\delta\cdot(1-\delta)^{n-1}$ $=C\cdot{\delta\over1-\delta} \sum_{n=1}^\infty \log(n) \cdot (1-\delta)^n$

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  • $\begingroup$ When you say "log" do you mean log base-10 or natural logarithm? $\endgroup$ – R. Burton Dec 9 '18 at 18:01
  • $\begingroup$ I suspect it has no closed form better than that. What do you need this for? $\endgroup$ – Ethan Bolker Dec 9 '18 at 18:01
  • $\begingroup$ Please edit the question to include a link to the Wolfram alpha output. $\endgroup$ – Shaun Dec 9 '18 at 18:11
  • $\begingroup$ @R.Burton natural logarithm $\endgroup$ – giuliolunati Dec 9 '18 at 19:19
  • $\begingroup$ @EthanBolker sorry, too long to explain now... maybe I'll edit the question. $\endgroup$ – giuliolunati Dec 9 '18 at 19:30
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Definition of Polylogarithm: http://mathworld.wolfram.com/Polylogarithm.html

No closed form exists in terms of elementary functions (addition, multiplication, powers, etc.), at least not in terms of real functions. You might be able to write it as a complex-valued function or improper integral.

Given that the polylogarithm is already a special function, I suspect that any closed form will be in terms of special functions rather than something nice.

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  • $\begingroup$ Yes, I read that, so I'd understand what $PolyLog(0,q)$ means. What I don't understand is the "exponent" $^{(1,0)}$ $\endgroup$ – giuliolunati Dec 9 '18 at 19:38
  • $\begingroup$ After working on it for a bit, I don't think that your sum has anything to do with polylogarithms at all. I have no idea what "$-PolyLog^{(1,0)}(0,x)$" is supposed to mean; it might be an error. Either way, The closest I can get to approximating the curve without spending more time on it is approximately $\frac{x^2}{1.2-x^2}$. $\endgroup$ – R. Burton Dec 9 '18 at 21:32
  • $\begingroup$ Thank you for spending time on that! Where that approximation come from? $\endgroup$ – giuliolunati Dec 9 '18 at 22:09
  • $\begingroup$ Wild guess based on the form of the graph (which is similar visually similar to to $\frac{x^2}{(a-x)^2}$, then adjusting $a$ until I got as close to $\sum_{n=1}^{1000 }\log(n)x^n$ as possible. $\endgroup$ – R. Burton Dec 9 '18 at 23:30
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Ok, I found here what the notation $f^{(1,0)}$ means in Wolfram Alpha: it's the derivative wrt the 1st variable.

So the response is $-\frac{\partial \operatorname{Li}(s,t)}{\partial s}\bigg|_{(s=0, t=x)}$

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