## Task 4 - 291234

Determine all positive real numbers $a$ for which the sequence $\left(x_{n}\right)$ defined by

$$
x_{1}=\sqrt{a}, \quad x_{n+1}=\sqrt{a+x_{n}} \quad(n=1,2, \ldots)
$$

is convergent; for each such number $a$, determine the limit of the sequence $\left(x_{n}\right)$.
We first show by complete induction that $x_{n+1}>x_{n}$ for all $n>1$:

Base case: it is $x_{2}=\sqrt{a+\sqrt{a}}>\sqrt{a}=x_{1}$.

Inductive step: for $x_{n+1}>x_{n}$ to hold, it must be true that:

$$
\sqrt{a+x_{n}}>x_{n}
$$

Since both sides are positive, we can square them:

$$
a+x_{n}>x_{n}^{2}
$$

or

$$
x_{n}^{2}-x_{n}-a<0
$$

This can be rewritten as:

$$
x_{n}^{2}-x_{n}-a-\frac{1}{4}<0
$$

Since by the problem statement $a>0$ is already given, there is no further restriction on $a$. The limit is obtained by solving the equation $x=\sqrt{x+a}$, and it is the only positive solution:

$$
x=\frac{1}{2}+\sqrt{\frac{1}{4}+a} \quad \forall \quad a>0
$$