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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)
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    $\begingroup$ 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 $\Re(x)>1$. $\endgroup$ – Steven Clark Apr 29 '20 at 17:39
  • $\begingroup$ @StevenClark Yeah, thanks for checking with Mathematica, because I don't use that at the moment. $\endgroup$ – Casimir Rönnlöf Apr 29 '20 at 18:05
  • $\begingroup$ You're welcome. By the way I think I should have written $\Re(x)\ge 1$ instead of $\Re(x)>1$. $\endgroup$ – Steven Clark Apr 29 '20 at 18:16
  • 1
    $\begingroup$ You can also simplify using Wolfram Alpha (see wolframalpha.com/input/…). $\endgroup$ – Steven Clark Apr 29 '20 at 18:29
  • $\begingroup$ @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. $\endgroup$ – Casimir Rönnlöf Apr 30 '20 at 4:11
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$H_{x-1}^{(n)}=\zeta(n)-\zeta(n,x)$ 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.
  2. I don't think there's anything new or found elsewhere. See item 4.
  3. Answered above.
  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.
  5. Who knows...
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