## SUBJECT 3.

Let the sequences $\left(x_{n}\right)_{n \geq 1},\left(y_{n}\right)_{n \geq 1}$ with $x_{1} \in(0,1)$ given, $x_{n+1}=x_{n}\left(1-x_{n}\right), \forall n \geq 1$ and $y_{n}=x_{1}^{2}+x_{2}^{2}+$ $\cdots+x_{n}^{2}$

a) Calculate $\lim _{n \rightarrow \infty} y_{n}$.

b) Calculate $\lim _{n \rightarrow \infty} n \cdot\left(x_{1}-y_{n}\right)$

Niculae Cavachi
## Subject 3.

a) $x_{k} \in(0,1), \forall k \geq 1$ (by induction)

$$
\begin{aligned}
& \frac{x_{k+1}}{x_{k}}=1-x_{k}\left(x_{n}\right) \downarrow=>\left(x_{n}\right)_{n} \text { convergent. } \\
& x_{n}^{2}=x_{n}-x_{n+1}, \forall n \geq 1 \\
& x_{1}^{2}=x_{1}-x_{2} \\
& x_{2}^{2}=x_{2}-x_{3} \stackrel{+}{\Rightarrow} y_{n}=x_{1}-x_{n+1}, \forall n \geq 1=>\lim _{n \rightarrow \infty} y_{n}=x_{1} \\
& x_{n}^{2}=x_{n}-x_{n+1}
\end{aligned}
$$

Let $l=\lim _{n \rightarrow \infty} x_{n}, l \in\left[0, x_{1}\right)=>l=l-l^{2}=>l=0$ thus $\lim _{n \rightarrow \infty} x_{n}=0$ ..... $\mathbf{1 p}$
b) $n \cdot\left(x_{1}-y_{n}\right)=\frac{n}{\frac{1}{x_{1}-y_{n}}}=\frac{a_{n}}{b_{n}}$, where $a_{n}=n, b_{n}=\frac{1}{x_{1}-y_{n}}, b_{n} \uparrow \infty$.

Then, according to the Stolz - Cesaro criterion, we have:

$$
\begin{aligned}
& \lim _{n \rightarrow \infty} n \cdot\left(x_{1}-y_{n}\right)=\lim _{n \rightarrow \infty} \frac{n+1-n}{\frac{1}{x_{1}-y_{n+1}}-\frac{1}{x_{1}-y_{n}}}=\lim _{n \rightarrow \infty} \frac{\left(x_{1}-y_{n}\right) \cdot\left(x_{1}-y_{n+1}\right)}{y_{n+1}-y_{n}}= \\
& =\lim _{n \rightarrow \infty} \frac{x_{n+1} \cdot x_{n+2}}{x_{n+1}^{2}}=\lim _{n \rightarrow \infty} \frac{x_{n+2}}{x_{n+1}}=\lim _{n \rightarrow \infty}\left(1-x_{n+1}\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\end{aligned}
$$