3. Calculate the approximate value of $\sqrt{41}$ using continued fractions.



Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. 

However, since the text provided is already in English, here is the equivalent text in the requested format:

3. Calculate the approximate value of $\sqrt{41}$ using continued fractions.
3. Solution: Since $6<\sqrt{41}<7$, we have
$$\begin{aligned}
\sqrt{41} & =6+(\sqrt{41}-6)=6+\frac{1}{\frac{1}{\sqrt{41}-6}} \\
& =6+\frac{1}{\frac{\sqrt{41}+6}{5}}
\end{aligned}$$

Since $2<\frac{\sqrt{41}+6}{5}<3$, we have
$$\begin{aligned}
\frac{\sqrt{41}+6}{5} & =2+\frac{\sqrt{41}-4}{5}=2+\frac{1}{\frac{5}{\sqrt{41}-4}} \\
& =2+\frac{1}{\frac{\sqrt{41}+4}{5}}
\end{aligned}$$

Since $2<\frac{\sqrt{41}+4}{5}<3$, we have
$$\begin{aligned}
\frac{\sqrt{41}+4}{5} & =2+\frac{\sqrt{41}-6}{5}=2+\frac{1}{\frac{5}{\sqrt{41}-6}} \\
& =2+\frac{1}{\sqrt{41}+6}
\end{aligned}$$

Since $12<\sqrt{41}+6<13$, we have
$$\begin{aligned}
\sqrt{41}+6 & =12+(\sqrt{41}-6) \\
& =12+\frac{1}{\frac{1}{\sqrt{41}-6}}=12+\frac{1}{\frac{\sqrt{41}+6}{5}}
\end{aligned}$$

The final fraction is the same as the last fraction in the first equation, so we get the periodic continued fraction representation of $\sqrt{41}$.
$$\begin{aligned}
\sqrt{41} & =6+\frac{1}{2+\frac{1}{2+\frac{1}{12+\frac{1}{2+\frac{1}{2+\frac{1}{12+}}}}}} \\
& =[6, \dot{2}, 2,12]
\end{aligned}$$

The first few convergents are
$$\begin{array}{l}
\frac{p_{1}}{q_{1}}=6, \quad \frac{p_{2}}{q_{2}}=\frac{13}{2}=6.5 \\
\frac{p_{3}}{q_{3}}=\frac{32}{5}=6.4, \quad \frac{p_{4}}{q_{4}}=\frac{397}{62}=6.403225 \cdots \\
\frac{p_{5}}{q_{5}}=\frac{826}{129}=6.403100 \cdots \\
\frac{p_{6}}{q_{6}}=\frac{2049}{320}=6.403125 \cdots
\end{array}$$

By Lemma 5, we have
$$6.403100<\sqrt{41}<6.403125$$