i equally thanked everybody once! PaulRS' solution is nice and my solution differs from his in the last setp only, where i applied Cesaro's mean, as Moo mentioned, to the sequence

$\displaystyle y_n=\frac{1}{x_{n+1}}-\frac{1}{x_n}.$ that will give you $\displaystyle \lim_{n\to\infty} n x_{n+1}=2$ and thus $\displaystyle \lim_{n\to\infty} (n+1)x_{n+1}=2.$