Evaluate the sum : $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\frac{1}{1+x_{3}}+...+\frac{1}{1+x_{n}}$

Question

Question :

Let the real number $x≥1$ :

$x_{1}=x$ and $x_{n+1}=x_{n}(1+x_{n})$ for

$n=1,2,3...$

Then find the sum :

$S=\displaystyle \sum_{k=1}^{n}\frac{1}{1+x_{k}}$

My try :

Note that : $\frac{1}{1+x_{1}}=\frac{1}{x_{1}}-\frac{1}{x_{2}}$

Also : $\frac{1}{1+x_{k}}=\frac{1}{x_{k}}-\frac{1}{x_{k+1}}$

So if we take sum rights-left we get :

$S=\frac{1}{x_{1}}-\frac{1}{x_{k+1}}$

I need see other method ?

And it's possible to find the term of $x_{n}$ ?