<s>[INST] Prove the positivity of a sequence Let $a>0$ a real number and $(u_n)$ the sequence defined by 
$$
u_{n+1} = a - \frac{1}{u_n}\text{ and } u_0 = a.
$$
Question: Determine condition on the value of $a>0$ such that the sequence $(u_n)$ is always positive. 
Attempt: I tried to establish a general formula of $u_n$ in order to set up a condition on $a$, by calculating $u_n$ with some $n$:
$$
u_1 = a - \frac{1}{u_0} = a - \frac 1a = \frac{a^2-1}{a},\\
u_2 = a - \frac{1}{u_1}= a - \frac{a}{a^2-1}=\frac{a^3-2a}{a^2-1},\\
u_3 = a - \frac{1}{u_2} = a - \frac{a^2-1}{a^3-2a}=\frac{a^4-3a^2+1}{a^3-2a},\\
u_4 = a - \frac{1}{u_3} = a - \frac{a^3-2a}{a^4-3a^2+1}=\frac{a^5-4a^3+2a-1}{a^4-3a^2+1}.
$$
But I wasn't successful, it seems that there is no general formula of $u_n$. So I would be appreciate for any suggestion of solution. Thank you!
 [/INST] Hint: $\;\require{cancel}
u_{n+1} - u_n = \left(\cancel{a} - \dfrac{1}{u_n}\right) -  \left(\cancel{a} - \dfrac{1}{u_{n-1}}\right)=\dfrac{u_n-u_{n-1}}{u_nu_{n-1}}
\,$, so as long as the terms are positive the differences between consecutive terms have the same sign i.e. the sequence is monotonic, and is in fact decreasing since $\,u_1-u_0=\left(\cancel{a} - \dfrac{1}{a}\right) - \cancel{a} \lt 0\,$. For it to be positive it is necessary and sufficient that $\,0\,$ is a lower bound, so the sequence is positive iff it is convergent. In this case, passing the recurrence relation to the limit, its limit $\,A\,$ must satisfy $\,A = a - \dfrac{1}{A}$ $\iff A^2 - aA + 1 = 0\,$. The latter equation has real roots iff $\,\Delta = a^2 - 4 \ge 0 \iff a \ge 2\,$.
</s>