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 $a=1/p$)

 A: $\newcommand{\+}{^{\dagger}}
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$\ds{\mbox{How does the closed form for}\
     \sum_{n = 1}^{\infty}\pars{{1 \over n} - {1 \over a + n}}\
     \mbox{look like ?}}$

\begin{align}
\!\!\sum_{n = 1}^{\infty}\pars{{1 \over n}\! - \!{1 \over a + n}}
&\!=a\sum_{n = 1}^{\infty}{1 \over n\pars{a + n}}
\!=a\sum_{n = 0}^{\infty}{1 \over \pars{n + 1 + a}\pars{n + 1}}
=\!a\,{\Psi\pars{1 + a} \!-\! \Psi\pars{1} \over \pars{1 + a} - 1}
\end{align}
  where $\ds{\Psi\pars{z}}$ is the Digamma Function $\ds{\bf\mbox{6.3.1}}$
  and we used identity
  $\ds{\bf\mbox{6.3.16}}$.

$$\color{#66f}{\large%
\sum_{n = 1}^{\infty}\pars{{1 \over n} - {1 \over a + n}}
=\Psi\pars{1 + a} + \gamma} = {\rm H}_{a}
$$

$\ds{\gamma = -\Psi\pars{1}}$ is the Euler-Mascheroni Constant $\ds{\bf\mbox{6.1.3}}$.
  $\quad\ds{\gamma \approx 0.5772156649\ldots\quad}$ $\ds{{\rm H}_{a}}$ is a generalized Harmonic Number.

A: 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.]
A: First, for any $z \in \mathbb{C}$ with $|z| < 1$, we have following expansion
$$-\log(1-z) = \sum_{n=1}^\infty \frac{z^n}{n}$$
Let $\displaystyle\;\omega = e^{\frac{i2\pi}{p}}\;$ be the primitive $p$-root of unity, we know for any integer $n$,
$$\frac{1}{p}\sum_{\ell=0}^{p-1} (\omega^\ell)^n = \delta_p(n) 
\stackrel{def}{=}
\begin{cases}1,& n \equiv 0\pmod p\\0,& \text{otherwise}\end{cases}$$
This implies for any integer $k \in \{\;1,2,\ldots,p\;\}$, we have
$$-\frac{1}{p}\sum_{\ell=0}^{p-1}\omega^{-k\ell}\log(1-z\omega^\ell)
= \frac{1}{p}\sum_{\ell=0}^{p-1}\sum_{n=1}^\infty \frac{z^n}{n}( \omega^\ell )^{n-k}
= \sum_{n=1}^\infty \delta_p(n-k)\frac{z^n}{n}
= \frac{1}{p}\sum_{n=0}^\infty \frac{z^{np+k}}{n+\frac{k}{p}}
$$
This leads to
$$\sum_{n=0}^\infty\left(\frac{1}{n+1} - \frac{1}{n+\frac{k}{p}}\right)z^{np}
= - \sum_{\ell=0}^{p-1}\left(z^{-p} - z^{-k}\omega^{-k\ell}\right)\log(1-z\omega^\ell)
\tag{*1}$$
Since the sequences $\displaystyle\;\frac{1}{n+1} - \frac{1}{n+\frac{k}{p}} \sim O\left(\frac{1}{n^2}\right)\;$, the series on the LHS converges absolutely as $z \to 1^{-}$. 
The RHS is a finite sum, the only term that may cause trouble is the term for $\ell = 0$.
It is clear the logarithm singularity from the $\log(1-z)$ get killed by the
$z^{-p} - z^{-k}$ factor as $z \to 1^{-}$. This allow us to take the limit $z \to 1^{-}$ in
$(*1)$ and get
$$\mathcal{S}_{k/p} \stackrel{def}{=}
\sum_{n=1}^\infty\left(\frac{1}{n} - \frac{1}{n+\frac{k}{p}}\right)
= \frac{p}{k} + \sum_{n=0}^\infty\left(\frac{1}{n+1} - \frac{1}{n+\frac{k}{p}}\right)\\
= \frac{p}{k} - \sum_{\ell=1}^{p-1}\left(1 - \omega^{-kl}\right)\log(1-\omega^\ell)
$$
Using the identities
$$\sum_{\ell=1}^{p-1}\log(1-\omega^\ell) = \log p
\quad\text{ and }\quad
\sum_{\ell=1}^{p-1}\omega^\ell = -1$$
The sum reduces to
$$\begin{align}
  &\frac{p}{k} - \log(2p)  + \sum_{l=1}^{p-1}\omega^{-k\ell}\log\left(\frac{1-\omega^\ell}{2}\right)\\
= & \frac{p}{k} - \log(2p) + \sum_{l=1}^{p-1}
\left( \cos\left(\frac{2\pi k\ell}{p}\right) - i\sin\left(\frac{2\pi k\ell}{p}\right)
\right)\left(\log\sin\left(\frac{\ell\pi}{p}\right) + i\pi\left(\frac{\ell}{p}-\frac12\right)\right)
\end{align}
$$
Using another set of trigonometric identities
$$\sum_{\ell=1}^{p-1}\frac{\ell}{p}\sin\left(\frac{2\pi k\ell}{p}\right) = -\frac12\cot\left(\frac{k\pi}{p}\right)
\quad\text{ and }\quad
\sum_{\ell=1}^{p-1}\sin\left(\frac{2\pi k\ell}{p}\right) = 0
$$
We finally get
$$
\bbox[4pt,border:1px solid blue]{
\mathcal{S}_{k/p} =
\frac{p}{k} - \log(2p) -\frac{\pi}{2}\cot\left(\frac{k\pi}{p}\right) + 
\sum_{l=1}^{p-1} 
\cos\left(\frac{2\pi k\ell}{p}\right) \log\sin\left(\frac{\ell\pi}{p}\right)
}
\tag{*2}
$$
On the wiki page of digamma function, there is a formula for digamma function at $\frac{k}{p}$.
$$\psi\left(\frac{k}{p}\right) = -\gamma - \log(2p) - \frac{\pi}{2}\cot\left(\frac{k\pi}{p}\right) + 2\sum_{\ell=1}^{\lfloor\frac{p-1}{2}\rfloor} \cos\left(\frac{2\pi k\ell}{p}\right)\log\sin\left(\frac{\ell\pi}{p}\right)$$
Compare this with what we have in $(*2)$, we can simplify our sum as
$$\mathcal{S}_{k/p} = \frac{p}{k}+\psi\left(\frac{k}{p}\right) + \gamma
= \psi\left(\frac{k}{p}+1\right) + \gamma$$
Replace $\frac{k}{p}$ by $a$, this matches the result derived in another answer.
