# Closed form for $\sum_{n=1}^\infty \log(n) * x^n$

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 $$r-\delta/2 . Then let's say that we can transmit $$r$$ via $$N$$ (with the tolerance $$\delta$$). So we need $$C\cdot\log(N)$$ bits...

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

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

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.

• Yes, I read that, so I'd understand what $PolyLog(0,q)$ means. What I don't understand is the "exponent" $^{(1,0)}$ – giuliolunati Dec 9 '18 at 19:38
• 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 $\frac{x^2}{1.2-x^2}$. – R. Burton Dec 9 '18 at 21:32
• Thank you for spending time on that! Where that approximation come from? – giuliolunati Dec 9 '18 at 22:09
• 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. – R. Burton Dec 9 '18 at 23:30

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