A series with logarithms 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?
 A: 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 $\mathcal{J}=\mathcal{S}$.
A: 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:
$$ S = \zeta(2)+\sum_{n\geq 1}|G_n|\int_{0}^{+\infty}\frac{x^n \log(x)}{1-x} \,dx = \boxed{\zeta(2)-\sum_{n\geq 1}^{\phantom{}}|G_n|\,\zeta(2,n+1)}.\tag{3}$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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$\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
$$
\,\mathcal{S} =
\ln\pars{2} + \sum_{n = 1}^{\infty}\bracks{%
2\,{\zeta\pars{2n} - 1 \over 2n - 1} - {\zeta\pars{n + 1} - 1 \over n}} =
-\ln\pars{2} + \sum_{n = 1}^{\infty}\bracks{%
{2\zeta\pars{2n} \over 2n - 1} - {\zeta\pars{n + 1} \over n}}
$$
A: 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
