calculus - 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.

Answer

$$\sum\limits_{n = 1}^\infty {\left( {\frac{1}{n} - \frac{1}{{n + x}}} \right)}=\sum\limits_{n = 1}^\infty \frac{x}{n(n+x)}$$

It is known that the sum is

$$\sum\limits_{n = 1}^\infty \frac{x}{n(n+x)}=\psi(x+1)+\gamma$$

where $\psi$ is the digamma function

$\gamma$ is the Euler constant .

Added :

We can easily prove that for $x=0.5$ we have

$$\psi(1.5)=2-\gamma -\log(4)$$

Hence the sum is equal to

$$2-\log(4)$$