Let's determine for every positive integer $n$ the first digit after the decimal point in the number $\sqrt{n^{2}+n+1}$.
When $n$ takes the values $1,2,3$, then $\sqrt{3}=1.7 \ldots ; \sqrt{7}=2.6 \ldots ; \sqrt{13}=3.6 \ldots$ We will show that if $n \geq 4$, then the digit in question is 5.

Since $n^{2}<n^{2}+n+1<(n+1)^{2}$, it follows that

$$
\left[\sqrt{n^{2}+n+1}\right]=n
$$

For the digit $\alpha$ after the decimal point, we have

$$
\begin{gathered}
n+\frac{\alpha}{10}<\sqrt{n^{2}+n+1}<n+\frac{\alpha+1}{10}, \quad \text { i.e., } \\
10 n+\alpha<10 \sqrt{n^{2}+n+1}<10 n+\alpha+1
\end{gathered}
$$

Squaring all terms of the inequality and subtracting $100 n^{2}$ from each term, we get

$$
20 n \alpha+\alpha^{2}<100 n+100<20 n(\alpha+1)+(\alpha+1)^{2}
$$

From the first inequality, after rearranging, we obtain

$$
20 n(\alpha-5)+\alpha^{2}<100
$$

The left side is increasing in both $n$ and $\alpha$, so if $n \geq 4$, then (1) can only hold if $\alpha \leq 5$.

From the second inequality, similarly,

$$
100<20 n(\alpha-4)+(\alpha+1)^{2} \leq 20 n(\alpha-4)+10^{2}, \quad \text { since } \quad \alpha \leq 9
$$

i.e.,

$$
0<20 n(\alpha-4)
$$

which is true if and only if $\alpha \geq 5$.

This proves that if $n$ is a positive integer greater than 3, then the digit in the decimal part of $\sqrt{n^{2}+n+1}$ is 5.