<s>[INST] How to calculate this expression? evaluate the expression [1]:
$$\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{n} - \frac{1}{{n + x}}} \right)} $$
where $x$ is a real number, $0\le x\le1$, and $x$ is rounded to 3 digits.
For example, when $x=0.500$, the expression is [2]:
$$\left(\frac11 -\frac1{1.5}\right)+\left(\frac12 -\frac1{2.5}\right)+\left(\frac13 -\frac 1{3.5}\right) + ...$$
For a given $x$, how can I evaluate it?
The answer must be rounded to more than 12 digits.
 [/INST] For $x=0$, the sum evaluates to $0$, and for $x=1$, it is $1$. 
For $0<x<1$,  you can derive a power series expression for the given sum as below:
\begin{align}
\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{x+n}\right)=&\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n}\left(1+\frac{x}{n}\right)^{-1}\right)\\
\ =& \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n}\left(1-\frac{x}{n}+\frac{x^2}{n^2}-\frac{x^3}{n^3}+\cdots\right)\right)\\
\ =&\sum_{n=1}^{\infty}\left(\frac{x}{n^2}-\frac{x^2}{n^3}+\frac{x^3}{n^4}-\cdots\right)\\
\ =&\zeta(2)x-\zeta(3)x^2+\zeta(4)x^3-\cdots\\
\ =&\sum_{k=1}^{\infty}\zeta(k+1)x^k
\end{align}
</s>