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?

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
