# Questions regarding $\ln(x) = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}(\zeta(n,x)-\zeta(n))$. Have I found something “new”?

Introduction

TL;DR I was messing around with the Taylor series for $$\ln(x)$$ when I ended up with the formula

\begin{align} \ln(x) &= \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}(\zeta(n,x)-\zeta(n)) \\\\ & =\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}H_{x-1}^{(n)}\end{align} (Here $$\zeta(n,x)$$ is Hurwit's Zeta function and $$H_{x-1}^{(n)}$$ is the $$(x-1)$$-th Harmonic number of order $$n$$ (Generalized Harmonic numbers))

I claim that this formula works for all $$x > 0$$ (only $$x\in\mathbb{R}$$ for now). My questions are at the bottom of the post.

Here are some numerical examples (using WolframAlpha):

Derivation

My derivation of the formula bases on the taylor series for $$\ln(x+1)$$ shown below

$${\displaystyle \ln(1+x)=\sum _{n=1}^{\infty}{\frac{(-1)^{n-1}}{n}}x^{n}}$$

which is valid for $$|x|\leq1$$. We can clearly see that we could get a infinite series for $$\ln(2)$$ by plugging in $$1$$. But how would we get a series for $$\ln(3)$$? Well, one could plug in $$\frac{1}{2}$$ to get that $${\displaystyle \ln(1+\frac{1}{2})=\sum _{n=1}^{\infty}{\frac{(-1)^{n-1}}{n2^n}}}$$ By adding the inside of the natural logarithm on the LHS, and then using basic logarithm properties we get: $${\displaystyle \ln(3)=\ln(2) + \sum _{n=1}^{\infty}{\frac{(-1)^{n-1}}{n2^n}}}$$

Then, using the infinite series from earlier for $$\ln(2)$$ we get

\begin{align} \ln(3) & =\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n2^n} \\\\ & = \sum_{n=1}^{\infty}\frac{(2^n+1)(-1)^{n+1}}{n2^n}\end{align}

Do you get the point? Now, in general, plugging in $$\frac{1}{x}$$, we would get:

\begin{align} \ln(x+1) & = \ln(x) + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{nx^n}\end{align}

Now what's $$\ln(x)$$? Well, one could do the exact same thing (the process I described above) for first $$x$$, then $$x-1$$, then $$x-2$$ and so on, all the way until $$1$$ since $$\ln(1) = 0$$. So doing this we get:

\begin{align} \ln(x+1) & = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n2^n} \cdots + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{nx^n} \\\\ & = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} + \frac{(-1)^{n+1}}{n2^n} \cdots + \frac{(-1)^{n+1}}{nx^n} \\\\ & = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\left(1+\frac{1}{2^n}+\frac{1}{3^n}\cdots+\frac{1}{x^n}\right) \\\\ & = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\sum_{k=1}^x \frac{1}{k^n}\\\\ & =\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}H_{x}^{(n)} \\\\ &= \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}(\zeta(n,x+1)-\zeta(n))\end{align}

Then plugging in $$x-1$$ we get: $$\boxed{\ln(x) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}H_{x-1}^{(n)} = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}(\zeta(n,x)-\zeta(n))}$$

Questions

1. First of all, is my derivation correct? (I believe so, since I have tested the formula numerically a lot now, and it has worked)
2. The title is a bit misleading; me finding something new about something elementary as natural logarithms is pretty much impossible, but I couldn't find this series listed anywhere, so if anyone recognizes this series please link some reference?
3. Does this series work for all $$x>0$$ and $$x\in\mathbb{R}$$? Maybe even complex numbers?
4. Does this series converge quickly?
5. Can something else be said about the series? (Cool things to note, possible simplifications... whatever)
• I'd say there's nothing new here as Mathematica simplifies $\sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} H_{x-1}^{(n)}$ to $\log(x)$, but it only seems to converge for $\Re(x)>1$. – Steven Clark Apr 29 '20 at 17:39
• @StevenClark Yeah, thanks for checking with Mathematica, because I don't use that at the moment. – Casimir Rönnlöf Apr 29 '20 at 18:05
• You're welcome. By the way I think I should have written $\Re(x)\ge 1$ instead of $\Re(x)>1$. – Steven Clark Apr 29 '20 at 18:16
• You can also simplify using Wolfram Alpha (see wolframalpha.com/input/…). – Steven Clark Apr 29 '20 at 18:29
• @StevenClark Yeah it should have been $\Re(x)$, no problem. I checked numerical examles with WolframAlpha and then I forgot to put the formula itself in...typical me. Anyways, thanks. – Casimir Rönnlöf Apr 30 '20 at 4:11

$$H_{x-1}^{(n)}=\zeta(n)-\zeta(n,x)$$ holds only if $$n>1$$. Anyway, we can use $$H_z^{(n)}=\sum_{k=1}^{\infty}\big(k^{-n}-(k+z)^{-n}\big),\qquad z\in\mathbb{C}\setminus\mathbb{Z}_{<0}.$$ With this, $$f(z):=\sum_{n=1}^{\infty}(-1)^{n-1}H_z^{(n)}/n$$ converges - to $$\ln(1+z)$$ - if and only if $$\color{blue}{|k+z|\geqslant 1}$$ for all positive integer values of $$k$$. For a proof, suppose $$z\notin\mathbb{Z}_{<0}$$, let $$K_1\subset\mathbb{Z}_{>0}$$ contain $$1$$ and all the (at most two) values of $$k$$ such that $$|z+k|\leqslant 1$$, and let $$K_2=\mathbb{Z}_{>0}\setminus K_1$$. Then, writing $$f(z)=f_1(z)+f_2(z),\qquad f_j(z)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\sum_{k\in K_j}\big(k^{-n}-(k+z)^{-n}\big),$$ we see that $$f_2(z)$$ converges absolutely, hence $$f(z)$$ converges if and only if $$f_1(z)$$ converges. This gives precisely the condition stated above (if there is a single value of $$k$$ with $$|k+z|<1$$, then $$(k+z)^{-n}$$ grows unbounded (in absolute value) with $$n$$; if there are two values, then these are $$k$$ and $$k+1$$ for some $$k$$, and then $$(k+z)^{-n}+(k+1+z)^{-n}$$ grows unbounded too). Assume it holds. The absolute convergence of $$f_2(z)$$ allows to switch the summations; as it's trivially allowed for $$f_1(z)$$ (since $$K_1$$ is finite), it's in fact allowed for the whole $$f(z)$$. Which gives \begin{align*} f(z)&=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\big(k^{-n}-(k+z)^{-n}\big) \\&=\sum_{k=1}^{\infty}\left[\ln\left(1+\frac{1}{k}\right)-\ln\left(1+\frac{1}{k+z}\right)\right] \\&=\lim_{n\to\infty}\sum_{k=1}^{n}\ln\frac{(k+1)(k+z)}{k(k+1+z)} \\&=\lim_{n\to\infty}\ln\frac{(n+1)(1+z)}{n+1+z}=\ln(1+z). \end{align*} Finally, here are my answers to the items of the question:
1. Yes, under the remark above about $$n=1$$, and the condition of $$x$$ being a positive integer.
4. I would say the opposite. $$H_z^{(n)}$$ doesn't go to $$0$$ as $$n\to\infty$$, so it is like $$\sum_{n=1}^{\infty}(-1)^{n-1}/n$$ itself.