<s>[INST] What is the closed form for $\sum_{n=1}^\infty \frac1n - \frac1{n+1/p}$? A while ago, I started to look at expressions of the following form:
$$
S_p:=\sum_{n=1}^\infty \frac1n - \frac1{n+1/p},
$$
 where $p$ is prime, because otherwise things get too complicated for me at the moment. 
What I found so far is the following:
$$
S_p= p -\log(2p) - \frac12 \pi \cot\left(\frac\pi p\right) + T_p,
$$
where $T_p$ contains $\frac{p-1}2$ terms $t_{p,k}$ of the form: 
\begin{cases}
\pm 2 \sin\left(\frac{k\pi}{2p}\right) \log\left(\sin\left(\frac{k\pi}{2p}\right)\right) \\
\pm 2 \cos\left(\frac{k\pi}{2p}\right) \log\left(\sin\left(\frac{k\pi}{2p}\right)\right)\\
\pm 2 \sin\left(\frac{k\pi}{2p}\right) \log\left(\cos\left(\frac{k\pi}{2p}\right)\right) 
\\
\pm 2 \cos\left(\frac{k\pi}{2p}\right) \log\left(\cos\left(\frac{k\pi}{2p}\right)\right) 
\end{cases} 
An example can be found here.

How does the closed form for $\sum_{n=1}^\infty \frac1n - \frac1{a+n}$ look like? (EDIT: where $a=1/p$)

 [/INST] Write your expression as the limit at $x \to 1$ of
$$\sum_{n = 1}^{\infty}  \frac{p x^{pn}}{pn} - \frac{p x^{pn+1}}{pn+1}.$$
If we differentiate this power series, we get
$$\sum_{n = 1}^{\infty} p x^{pn - 1} - p x^{pn} 
=
p x^{p-1} \frac{1-x}{1 - x^p}. 
$$
Integrating this (say via a partial fraction decomposition)
and then choosing the constant of integration so that you get
zero when $x = 0$ gives a formula for the original power series.
Now putting $x = 1$ gives the value you want.
E.g. if $p = 2$, we have
$$2 x \frac{1}{1+x} =  2 - \dfrac{2}{1+x},$$
whose desired integral is 
$$2x - 2\log (1+x),$$
and so the value of the sum in this case is $2-2\log 2.$
E.g. if $p = 3$, we have
$$3 \frac{x^2}{1+x+x^2} = 3 - 3\frac{1+x}{1+x+x^2} = 3 - \frac{3}{2} \frac{2x+1}{1+x+x^2} - \frac{3}{2}\frac{1}{3/4+ (x+1/2)^2},$$
whose desired integral is
$$3 x - \frac{3}{2}\log(1+x+x^2) - {\sqrt{3}}\arctan\bigl(\frac{2x+1}{\sqrt{3}}\bigr)+\sqrt{3} \frac{\pi}{6},$$
and so the value of the sum in this case is
$$3 - \frac{3\log 3}{2} - \frac{\sqrt{3} \pi}{6}.$$
I didn't try to find the general formula here.  One part of mathematics where it is
worked out is the theory of special values (at $s = 1$) of Dirichlet $L$-series.  
[Added: it seems that between when I began writing this yesterday, and when I finally got a chance to finish and post it now, other answers with related derivations have appeared. Oh well.]
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