$$ a_{n+1}=(a_1+a_2+\cdots+a_{n+1})-(a_1+a_2+\cdots+a_n)= 2-\dfrac{1}{x_{n+1}}-2+\frac{1}{x_n}=\frac{1}{x_n}-\frac{1}{x_{n+1}} $$Answer is $x_n=n \ \forall \ n\in Z^+$. Since $\frac{1}{x_{n-1}}-\frac{1}{x_n}=a_n>0$ we have $x_n>x_{n-1}$. Also $x_5=5$ hence $x_1=1, \ x_2=2, \ x_3=3, \ x_4=4, \ x_5=5$. \[\frac{1}{4}-\frac{1}{5}\geq a_6=\frac{1}{5}-\frac{1}{x_6}\implies \frac{1}{x_6}\geq \frac{3}{20}\implies x_6\leq 6\]Thus, $x_6=6$. Now we will induct on $n$ with $x_n=n, \forall n\leq N$ and we will show that $x_{N+1}=N+1$. Our base cases are $n\leq 6$. \[\frac{1}{N^2-N}=\frac{1}{N-1}-\frac{1}{N}\geq a_{N+1}=\frac{1}{N}-\frac{1}{x_{N+1}}\]Which yields \[\frac{1}{x_{N+1}}\geq \frac{1}{N}-\frac{1}{N^2-N}=\frac{N-2}{N^2-N}>\frac{N-2}{N^2-4}=\frac{1}{N+2}\]This gives that $N+2>x_{N+1}>N\iff x_{N+1}=N+1$ as desired.$\blacksquare$