If $x_{n+1}= \frac{x_n^2+x_n+1}{x_n+1}$ find$ \sum_{n=1}^{p}\frac{1}{1+x_n}$ If $x_{n+1}= \frac{x_n^2+x_n+1}{x_n+1}$ find $ \sum_{n=1}^{p}\frac{1}{1+x_n}$, 
given $x_1=1/2$
The original question was for p=2014, but I want to find the general solution. 
I have tried using $x^2+x+1=(x- \omega)(x-\omega^2)$ and breaking $1/x_{n+1}$ into factors, but yielded no results.
I tried expressing $\frac{1}{1+x_{n+1}}$ in terms of $x_n$ but still no result.
Any help/hints on how to proceed will be highly appreciated.

Observe that $x_{n + 1} = x_n + \frac {1} {1 + x_n}$ so $\frac {1} {1 + x_n} = x_{n + 1} - x_n$ so $$\sum_{n = 1}^{p} \frac {1} {1 + x_n} = \sum_{n = 1}^{p} (x_{n + 1} - x_n) = x_{p + 1} - x_1 = x_{p + 1} - \frac {1} {2}.$$
