11.8. Let the sequence $\left(a_{n}\right)_{n=1}^{\infty}, a_{n}=\underbrace{11 \ldots 1}_{n \text { digits }} \underbrace{88 \ldots 8}_{\text {digits }}, \forall n \geq 1$. Calculate $\lim _{n \rightarrow \infty}\left\{\sqrt{a_{n}}\right\}$, where $\{t\}$ represents the fractional part of the number $t$.
## Solution.

Let $b_{n}=\underbrace{11 \ldots 1}_{n \text { digits }}$ and $c_{n}=3 b_{n}, \forall n \geq 1$. We observe that $\lim _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} c_{n}=+\infty$. We have

$$
\begin{aligned}
a_{n} & =\underbrace{11 \ldots 1}_{n \text { digits }} \underbrace{188}_{\text {n digits }}=\underbrace{11 \ldots 1}_{n \text { digits }} \underbrace{00 \ldots 0}_{n \text { digits }}+\underbrace{88 \ldots 8}_{\text {n digits }}=b_{n} \cdot 10^{n}+8 \cdot b_{n}=b_{n}\left(10^{n}+8\right)= \\
& =b_{n}\left(10^{n}-1+9\right)=b_{n}\left(9 b_{n}+9\right)=3 b_{n}\left(3 b_{n}+3\right)=c_{n}\left(c_{n}+3\right), \quad \forall n \geq 1
\end{aligned}
$$

from which it follows that $\left(\sqrt{a_{n}}\right)^{2}=a_{n}=c_{n}\left(c_{n}+3\right)=c_{n}^{2}+3 c_{n}c_{n}^{2}+2 c_{n}+1=\left(c_{n}+1\right)^{2}, \forall n \geq 1$, i.e., $c_{n}+1<\sqrt{a_{n}}<c_{n}+2, \forall n \geq 1$, which implies $\left[\sqrt{a_{n}}\right]=c_{n}+1, \forall n \geq 1$. We obtain

$$
\begin{aligned}
& \left\{\sqrt{a_{n}}\right\}=\sqrt{a_{n}}-\left[\sqrt{a_{n}}\right]=\sqrt{c_{n}\left(c_{n}+3\right)}-\left(c_{n}+1\right)=\frac{\left(\sqrt{c_{n}\left(c_{n}+3\right)}-\left(c_{n}+1\right)\right)\left(\sqrt{c_{n}\left(c_{n}+3\right)}+\left(c_{n}+1\right)\right)}{\sqrt{c_{n}\left(c_{n}+3\right)}+\left(c_{n}+1\right)}= \\
& =\frac{c_{n}\left(c_{n}+3\right)-\left(c_{n}+1\right)^{2}}{\sqrt{c_{n}\left(c_{n}+3\right)}+\left(c_{n}+1\right)}=\frac{c_{n}-1}{\sqrt{c_{n}\left(c_{n}+3\right)}+\left(c_{n}+1\right)}=\frac{1-\frac{1}{c_{n}}}{\sqrt{1+\frac{3}{c_{n}}}+1+\frac{1}{c_{n}}} \rightarrow \frac{1-0}{\sqrt{1+0}+1+0}=\frac{1}{2}(n \rightarrow \infty) .
\end{aligned}
$$

Thus, $\lim _{n \rightarrow \infty}\left\{\sqrt{a_{n}}\right\}=\frac{1}{2}$.