## University Calculus: Early Transcendentals (3rd Edition)

$\{x_{n}\}=1,3,7,17,39,...$ $\{y_{n}\}=1,2,5,12,17,...$ $a.\quad$ Verify $x_{1}^{2}-2y_{1}^{2}=-1,\qquad x_{2}^{2}-2y^{2}=+1$ $1^{2}-2(1^{2})=1-2=-1$ $3^{2}-2(2^{2})=9-8=+1$ (verified) Verify that $(a+2b)^{2}-2(a+b)^{2}=\pm 1,\ \quad$if $a^{2}-2b^{2}=\pm 1$ $a^{2}+4ab+4b^{2}-2(a^{2}+2ab+b^{2})=$ $=a^{2}+4ab+4b^{2}-2a^{2}-4ab-2b^{2}$ $=-a^{2}+2b^{2}$ $=-(a^{2}-2b^{2})\qquad$ (by assumption, the parentheses are either 1 or -1 $=-(\pm 1)$ $=\mp 1\qquad$ (which is either 1 or -1) Hence, verified. $b.\quad$ \begin{align*} r_{n}^{2}-2&=[\displaystyle \frac{a+2b}{a+b}]^{2}-2 & & \\ r_{n}^{2}-2& =\displaystyle \frac{ (a+2b)^{2}-2(a+b)^{2}}{(a+b)^{2}} &\text{ ... use the result of a.} & \\ r_{n}^{2}-2&=\displaystyle \frac{ -(a^{2}-2b^{2})}{y_{n}^{2}} &\text{ ... use the result of a.} & \\ r_{n}^{2}&=2+\displaystyle \frac{ \pm 1}{y_{n}^{2}} & \\ r_{n}&=\sqrt{2+\frac{ \pm 1}{y_{n}^{2}}} & \end{align*}. As n grows very large, it seems that $y_{n}$ also grows very large, $\{y_{n}\}=1,2,5,12,17,...$ We want to prove that from the third term onward, $y_{n}\geq n$. Proof by induction: Let $n=3$. The third denominator is $5$, so $y_{n}\geq n$, for $n=3$ (base step of induction) (Induction step) Assume that the the nth denominator is $\geq n.$ Let $r_{n}=\displaystyle \frac{a}{b}, \quad (x_{n}=a,y_{n}=b\geq n$ and $a\geq b)$ Then , the (n+1)st denominator $y_{n}$ equals $(a+b)\geq(b+b)=2b$ $y_{n+1}\geq 2b \qquad$(b is the nth denominator) $y_{n+1}\geq 2n\qquad$(by the induction step hypothesis) $y_{n+1}\geq n+1 \qquad$(2n $\geq$ n+1 for $n\geq 3)$ which proves the statement. Thus, $\displaystyle \frac{ \pm 1}{y_{n}^{2}}\rightarrow 0$, when $n\rightarrow\infty$. And, we have that $\displaystyle \lim_{n\rightarrow\infty}r_{n}=\sqrt{2}$