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?

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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
$$
\,\mathcal{S} =
\ln\pars{2} + \sum_{n = 1}^{\infty}\bracks{%
2\,{\zeta\pars{2n} - 1 \over 2n - 1} - {\zeta\pars{n + 1} - 1 \over n}} =
-\ln\pars{2} + \sum_{n = 1}^{\infty}\bracks{%
{2\zeta\pars{2n} \over 2n - 1} - {\zeta\pars{n + 1} \over n}}
$$
