A series with logarithms

Question

Can we express in terms of known constants the sum:

$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{\log (n+1)-\log n}{n}$$

First of all it converges , but not matter what I try or whatever technic I am about to apply it fails. In the mean time if we split it apart (let us take the partial sums) then:

$$\sum_{n=1}^{N} \frac{\log (n+1) - \log n}{n}= \sum_{n=1}^{N} \frac{\log (n+1)}{n} - \sum_{n=1}^{N} \frac{\log n}{n}$$

The last sum at the RHS does resemble a zeta function derivative taken at $1$. Of course zeta function diverges at $1$ but its PV exists, namely $\mathcal{P}(\zeta(1))=\gamma$. Maybe we have a PV for the derivative also? The other sum at the RHS is nearly the last sum at the right.

This is as much as I have noticed in this sum. Any help?

Addendum: I was trying to evaluate the integral:

$$\mathcal{J}=\int_0^1 \frac{(1-x) \log (1-x)}{x \log x} \, {\rm d}x$$

This is what I got.

\begin{align*} \int_{0}^{1}\frac{(1-x) \log(1-x)}{x \log x} &=-\int_{0}^{1} \frac{1-x}{x \log x} \sum_{n=1}^{\infty} \frac{x^n}{n} \, {\rm d}x \\ &= -\sum_{n=1}^{\infty}\frac{1}{n} \int_{0}^{1}\frac{x^{n-1} (1-x)}{\log x} \, {\rm d}x\\ &=\sum_{n=1}^{\infty} \frac{1}{n} \int_0^1 \frac{x^n-x^{n-1}}{\log x} \, {\rm d}x \\ &\overset{(*)}{=} \sum_{n=1}^{\infty} \frac{\log(n+1) -\log n}{n} \\ &= ? \end{align*}

$(*)$ since it is quite easy to see that:

$$\int_{0}^{1}\frac{x^a-x^{a-1}}{\log x} \, {\rm d}x = \log (a+1) - \log a , \; a \geq 1$$

due to DUTIES.

Maybe someone else can tackle the integral in a different manner?

Perhaps progress might be made by rearranging the negative terms to associate to the right instead, which gives $ \sum_{n=2}^\infty \frac{\log n}{n(n-1)} $? — hmakholm left over Monica, Jul 20 '16 at 14:24
Hi nospoon.. For the evaluation of the integral you mention maybe use the fact that Harmonic number is related to digamma ? — Tolaso, Jul 20 '16 at 14:48
The series is equivalent to $$\sum_{n=1}^{\infty}\frac{\ln(n+1)}{n(n+1)}$$ If that helps anyone. — Will Fisher, Jul 20 '16 at 15:18
@Tolaso Pllease there is a typo in your question, it's rather $\mathcal{J}=\mathcal{S}$, not $\mathcal{J}=-\mathcal{S}$. — Olivier Oloa, Jul 20 '16 at 15:22
@WillFisher Yes, this form is noticed in Theorem $2$ here: http://math.stackexchange.com/questions/364452/evaluate-int-0-frac-pi2-frac11x21-tan-x-mathrm-dx/1381569#1381569 — Olivier Oloa, Jul 20 '16 at 15:36

score 3 · Accepted Answer · edited Apr 13 '17 at 12:21

The given series admits a closed-form in terms of the poly-Stieltjes constants. The poly-Stieltjes constants arise in the context of finding the Laurent series expansion of the poly-Hurwitz zeta function $$ \begin{align} \zeta(s\mid a,b)= \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, \tag1 \end{align} $$ around $s = 0$. One may prove that (see Theorem $1$), as $s \to 0$, $$\zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{+\infty} \frac{(-1)^{k}}{k!}\gamma_k(a,b) s^k,\tag2 $$ with $$ \begin{align} \gamma_k(a,b)& = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log^k (n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right),\tag3 \\\\ \gamma_{k}(a,a)&=\gamma_{k}(a+1),\tag4 \end{align} $$ where $\gamma_{k}(a+1)$ are the generalized Stieltjes constants.

We thus obtain the following result.

Proposition. $$ \begin{align} &\mathcal{S}=\sum_{n=1}^{\infty} \frac{\log (n+1)-\log n}{n}=\gamma_1(1,0)-\gamma_1 \tag5 \\\\&\mathcal{J}=\int_0^1 \frac{(1-x) \log (1-x)}{x \log x} \, {\rm d}x=\gamma_1(1,0)-\gamma_1 \tag6 \end{align} $$

by using Theorem $2$ and the fact that $\mathcal{J}=\mathcal{S}$.

score 2 · Answer 2 · answered Jul 20 '16 at 15:58

We may exploit Frullani's theorem to get an integral representation of our series.

$$\begin{eqnarray*}S=\sum_{n\geq 1}\frac{\log(n+1)-\log(n)}{n}&=&\int_{0}^{+\infty}\sum_{n\geq 1}\frac{e^{-nx}-e^{-(n+1)x}}{nx}\,dx\\ &=&\int_{0}^{+\infty}\frac{1-e^{-x}}{x}\left(-\log(1-e^{-x})\right)\,dx\\&=&\int_{0}^{1}\frac{x\log x}{(1-x)\log(1-x)}\,dx\tag{1}\end{eqnarray*}$$ In terms of Gregory coefficients $$ \frac{x}{\log(1-x)}=-1+\sum_{n\geq 1}|G_n|x^n\tag{2}$$ gives: $$ S = \zeta(2)+\sum_{n\geq 1}|G_n|\int_{0}^{+\infty}\frac{x^n \log(x)}{1-x} \,dx = \boxed{\zeta(2)-\sum_{n\geq 1}^{\phantom{}}|G_n|\,\zeta(2,n+1)}.\tag{3}$$

+1. Nice job. You're a Frullani fan. The series looks like an innocent one whatsoever but... — Felix Marin, Jul 21 '16 at 04:44

Felix Marin · Answer 3 · 2016-07-21T22:14:28.427

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\,\mathcal{S} = \sum_{n = 1}^{\infty}{\ln\pars{n + 1} - \ln\pars{n} \over n} =\, ?}$.

\begin{align} \color{#f00}{\,\mathcal{S}} & = \sum_{n = 1}^{\infty}{\ln\pars{n + 1} - \ln\pars{n} \over n} = \ln\pars{2} + \sum_{n = 2}^{\infty}{\ln\pars{n + 1} - \ln\pars{n} \over n} \\[4mm] & = \ln\pars{2} + \sum_{n = 2}^{\infty}{1 \over n}\,\ln\pars{n + 1 \over n - 1} - \sum_{n = 2}^{\infty}{1 \over n}\ln\pars{n \over n - 1}\tag{1} \end{align}

The second term: \begin{equation} c \equiv \sum_{n = 2}^{\infty}{1 \over n}\ln\pars{n \over n - 1} = \sum_{n = 1}^{\infty} {\zeta\pars{n + 1} - 1 \over n}\tag{2} \end{equation} is related to the Alladi-Grinstead Constant $\ds{\expo{c - 1} = 0.809394020540639\ldots}$ while the sum $\ds{\sum_{n = 2}^{\infty} {1 \over n}\,\ln\pars{n + 1 \over n - 1}}$ still 'claims' for a 'closed form'. Similarly, by expanding $\ds{\ln\pars{n \over n - 1}}$ in powers of $\ds{n^{-1}}$, $\pars{1}$ becomes $$ \,\mathcal{S} = \ln\pars{2} + \sum_{n = 1}^{\infty}\bracks{% 2\,{\zeta\pars{2n} - 1 \over 2n - 1} - {\zeta\pars{n + 1} - 1 \over n}} = -\ln\pars{2} + \sum_{n = 1}^{\infty}\bracks{% {2\zeta\pars{2n} \over 2n - 1} - {\zeta\pars{n + 1} \over n}} $$

Felix Marin: I believe that this:
$$c \equiv \sum_{n = 2}^{\infty}{1 \over n}\ln\pars{n \over n - 1} = \sum_{n = 1}^{\infty} {\zeta\pars{n} - 1 \over n}\tag{2}$$

should be

$$c \equiv \sum_{n = 2}^{\infty}{1 \over n}\ln\pars{n \over n - 1} = \sum_{n = 1}^{\infty} {\zeta\pars{n+1} - 1 \over n}\tag{2}$$

No? — Tolaso, Jul 21 '16 at 18:54
.... because otherwise we have:
$$\sum_{n=2}^{\infty} \frac{\zeta(n)-1}{n}=1-\gamma$$ — Tolaso, Jul 21 '16 at 18:55

score 1 · Answer 4 · edited Apr 13 '17 at 12:20

Just an addendum, maybe some of this will be useful to you:

$$\int_0^1 \frac{(1-x) \log (1-x)}{x \log x} \, {\rm d}x=$$

$$=1.2577468869\dots= \gamma_{1}(1,0) - \gamma=\int_0^1 \ln (1-x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx-\gamma=$$

$$=\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\sum_{k = 2}^{\infty} \frac{(-1)^k \zeta(k)}{k-1}=$$

$$= \frac{\pi^2}{4}-1-4\int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1}=\frac{\pi^2}{4}-1-4\int_{0}^{\pi/2} \frac{t~dt}{e^{\pi \tan t}+1} =$$

$$=\int_{1}^{\infty} \frac{\ln ([x]+1)}{x^2}dx=\int^{\infty}_0 \frac{\log x+\Gamma (0,x) + \gamma}{e^x-1}~dx$$

Among the sources of the above expressions are:

http://www.people.fas.harvard.edu/~sfinch/csolve/kz3.pdf

http://www.people.fas.harvard.edu/~sfinch/csolve/kz.pdf

https://math.dartmouth.edu/~carlp/factorial.pdf

https://books.google.com/books?id=Pl5I2ZSI6uAC&pg=PA122&lpg=PA122&dq=1.2577468869&source=bl

https://math.stackexchange.com/a/1065075/269624

https://math.stackexchange.com/a/1733543/269624

A series with logarithms

4 Answers4