Theorem 1   Let $d$ be a positive integer which is not a perfect square, and let $(x,y)$ be a solution of $x^2 -dy^2 =1$ Then $\frac{x}{y}$ is a convergent in the simple continued fraction expansion of $\sqrt{d}$

Proof. Suppose we have a non-trivial solution $x,y$ of Pell's equation, i.e. $y \neq 0$ Let $x,y$ both be positive integers. From $$\left(\frac{x}{y}\right)^2 =d +\frac{1}{y^2}$$ we see that $\left(\frac{x}{y}\right)^2 > d$ hence $\frac{x}{y} > \sqrt{d}$ So we get \begin{eqnarray*} \left\vert \frac{x}{y} -\sqrt{d}\right\vert =\frac{1}{y^2\left(\frac{x}{y} +\sqrt{d}\right)} & < \frac{1}{y^2\left(2\sqrt{d}\right)} \\ & < \frac{1}{2y^2}. \end{eqnarray*}This implies that $\frac{x}{y}$ is a convergent of the continued fraction of $\sqrt{d}$