## Task 6B - 241246B

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

$$
x_{1}=1, \quad x_{n+1}=\sqrt{k\left(x_{n}^{2}+x_{n}\right)} \quad(n=1,2,3, \ldots)
$$

is convergent.

For each such number $k$, determine the limit of the sequence $\left(x_{n}\right)$.
For $k=1$, $x_{n+1}^{2}=x_{n}^{2}+x_{n} \geq x_{n}^{2}+1 \geq n+1$, which can be easily proven by induction. In particular, the sequence $\left(x_{n}^{2}\right)$ definitely diverges to $+\infty$, and thus the sequence $\left(x_{n}\right)$ as well.

For larger values of $k$, the terms of the sequence increase further due to the monotonicity of the recursion condition for calculating $x_{n+1}$ from $k$ and $x_{n}$, which can also be shown simply by induction, so that the sequences also diverge to $+\infty$ in this case.

For $0 < k < 1$, the sequence is bounded and converges to a limit $g$. The function $f(x) = x^2 + kx$ is strictly increasing for $x > 0$ and $0 < k < 1$. Therefore, the sequence is monotonically increasing and bounded above by $\frac{k}{1-k}$. If the sequence converges to a limit $g$, then $g$ must satisfy the equation $g = g^2 + kg$. Solving this equation, we get $g = \frac{k}{1-k}$, which is the only positive solution. 

In the case $\frac{1}{2} \leq k < 1$, the sequence is also bounded above by $\frac{k}{1-k}$ and converges to $g = \frac{k}{1-k}$. For $0 < k < \frac{1}{2}$, the sequence is bounded below by $0$ and converges to $g = \frac{k}{1-k}$.

In summary, the sequence diverges to $+\infty$ for $k \geq 1$ and converges to $g = \frac{k}{1-k}$ for $0 < k < 1$.