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Define $({a}_{n})$ inductively by putting ${a}_{1}={a}_{2}=1$ and ${a}_{n+2}={a}_{n}+{a}_{n+1}$, $\forall n\in N$. Write ${x}_{n}=\frac{{a}_{n}}{{a}_{n+1}}$ and prove that $\lim {x}_{n}=\phi$, where $\phi$ is the only positive real such that $\frac{1}{1+\phi}=\phi$.

The hint is to define $f\left(n+2 \right)=f\left(n+1 \right)+f\left(n \right)$ with initial conditions $f\left(1 \right)=f\left(2 \right)=1$, defining ${x}_{n}=\frac{f\left(n \right)}{f\left(n+1 \right)}$ then $\lim {x}_{n}=\phi$ positive root of $x^2 + x - 1=0$.

I do not understand how one concludes that this $\phi$ is root

Shatrix
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1 Answers1

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Note that $x_n = \frac{a_n}{a_{n+1}} = \frac{a_{n}}{a_{n} + a_{n-1}} = \frac{a_{n-1}+a_{n} - a_{n-1}}{a_{n}+a_{n-1}} = 1 - \frac{a_{n-1}}{a_n + a_{n-1}} = 1 - \dfrac{\dfrac{a_{n-1}}{a_n}}{\dfrac{a_n + a_{n-1}}{a_n}} = 1 - \frac{x_{n-1}}{1+{x_{n-1}}}$.

Simplifying this, we get $x_n = \frac{1}{x_{n-1}+1}$.

Now, note that $x_n$ is bounded. To show this, see that $x_n \leq 1$ for all $n$, since $a_n$ is increasing. Furthermore, $x_n \geq \frac{1}{2}$ for all $n \geq 1$, since $\frac{a_{n}}{a_{n+1}} = \frac{a_n}{a_n + a_{n-1}} \geq \frac{a_n}{a_n + a_n} \geq \frac 12 $.

I leave you to show that beyond $n>2$, $x_n$ is increasing (use induction). Hence, $\lim x_n$ exists as it is monotone bounded.

Now, using the equation for $x_n$ and $x_{n-1}$, we see that if $l = \lim x_n$, then $l = \frac{1}{l+1}$, just taking limit as $n \to \infty$ on both sides. This gives us the result that we want.

Sarvesh Ravichandran Iyer
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