25.43. Let $(1+\sqrt{2}+\sqrt{3})^{n}=q_{n}+r_{n} \sqrt{2}+s_{n} \sqrt{3}+t_{n} \sqrt{6}$, where $q_{n}, r_{n}, s_{n}, t_{n}$ are natural numbers. Compute the limits $\lim _{n \rightarrow \infty} \frac{r_{n}}{q_{n}}, \lim _{n \rightarrow \infty} \frac{s_{n}}{q_{n}}$ and $\lim _{n \rightarrow \infty} \frac{t_{n}}{q_{n}}$.

## 25.6. Least Upper Bound

Let $X$ be some (non-empty) set of real numbers. A number $x_{0}$ is called the least upper bound of this set if $x \leqslant x_{0}$ for any number $x$ in the set $X$, but for

## 300  Chapter 25. Limit of a Sequence

any $\varepsilon>0$ there exists a number $x$ in the set $X$ such that $x+\varepsilon>x_{0}$. If the least upper bound $x_{0}$ of the set $X$ belongs to $X$, then $x_{0}$ is called the maximum of the set $X$. Similarly, the greatest lower bound and the minimum are defined.
25.43. Let $\lambda_{1}=1+\sqrt{2}+\sqrt{3}, \lambda_{2}=1-\sqrt{2}+\sqrt{3}, \lambda_{3}=1+\sqrt{2}-\sqrt{3}$, $\lambda_{4}=1-\sqrt{2}-\sqrt{3}$. Then

$$
\begin{aligned}
& \lambda_{1}^{n}=q_{n}+r_{n} \sqrt{2}+s_{n} \sqrt{3}+t_{n} \sqrt{6}, \\
& \lambda_{2}^{n}=q_{n}-r_{n} \sqrt{2}+s_{n} \sqrt{3}-t_{n} \sqrt{6}, \\
& \lambda_{3}^{n}=q_{n}+r_{n} \sqrt{2}-s_{n} \sqrt{3}-t_{n} \sqrt{6}, \\
& \lambda_{4}^{n}=q_{n}-r_{n} \sqrt{2}-s_{n} \sqrt{3}+t_{n} \sqrt{6} .
\end{aligned}
$$

Therefore,

$$
\begin{aligned}
& \lambda_{1}^{n}+\lambda_{2}^{n}+\lambda_{3}^{n}+\lambda_{4}^{n}=4 q_{n}, \\
& \lambda_{1}^{n}+\lambda_{2}^{n}-\lambda_{3}^{n}-\lambda_{4}^{n}=4 s_{n} \sqrt{3}, \\
& \lambda_{1}^{n}-\lambda_{2}^{n}+\lambda_{3}^{n}-\lambda_{4}^{n}=4 r_{n} \sqrt{2}, \\
& \lambda_{1}^{n}-\lambda_{2}^{n}-\lambda_{3}^{n}+\lambda_{4}^{n}=4 t_{n} \sqrt{6} .
\end{aligned}
$$

It is also clear that $\lim _{n \rightarrow \infty} \frac{\lambda_{2}^{n}}{\lambda_{1}^{n}}=\lim _{n \rightarrow \infty} \frac{\lambda_{3}^{n}}{\lambda_{1}^{n}}=\lim _{n \rightarrow \infty} \frac{\lambda_{4}^{n}}{\lambda_{1}^{n}}=0$. Therefore, $\lim _{n \rightarrow \infty} \frac{\lambda_{1}^{n}}{q_{n}}=4$, $\lim _{n \rightarrow \infty} \frac{r_{n}}{\lambda_{1}^{n}}=\frac{1}{4 \sqrt{2}}, \lim _{n \rightarrow \infty} \frac{s_{n}}{\lambda_{1}^{n}}=\frac{1}{4 \sqrt{3}}, \lim _{n \rightarrow \infty} \frac{t_{n}}{\lambda_{1}^{n}}=\frac{1}{4 \sqrt{6}}$, and thus, $\lim _{n \rightarrow \infty} \frac{r_{n}}{q_{n}}=\frac{1}{\sqrt{2}}$, $\lim _{n \rightarrow \infty} \frac{s_{n}}{q_{n}}=\frac{1}{\sqrt{3}}$ and $\lim _{n \rightarrow \infty} \frac{t_{n}}{q_{n}}=\frac{1}{\sqrt{6}}$.