What is the limit of the sequence {$a_n$}?

[lfdahl]

Let the sequence {$a_n$} be defined by $a_1 = 1$ and $a_n = \frac{1}{1+a_{n-1}}$

for $n \ge 2$ . Find the limit.

 

[chisigma]

Let the sequence {$a_n$} be defined by $a_1 = 1$ and $a_n = \frac{1}{1+a_{n-1}}$ for $n \ge 2$ . Find the limit.

[sp]The difference equation can be written as... $\displaystyle \Delta_{n} = a_{n=1} - a_{n} = \frac {1}{1 + a{n}} - a_{n}= \frac{1 - a_{n} - a^{2}_{n}}{1 + a_{n}}= f(a_{n})\ (1)$ ... and there is the only attractive fixed point of f(x) in $\displaystyle \xi = \frac {\sqrt{5}-1}{2} = .618...$, so that for any $x_{0}> - 1$ he sequence tends to $\xi$...[/sp] Kind regards $\chi$ $\sigma$

 

[lfdahl]

Thankyou, chisigma, for your valuable and correct contribution!
The thread is still open for alternative approaches.

 

[Saitama]

Spoiler

Substitute $a_n$ repeatedly i.e
$$a_n=\frac{1}{1+a_{n-1}}=\frac{1}{1+\frac{1}{1+a_{n-2}}}=\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}}$$
Since $n\rightarrow \infty$, we can write:
$$a_n=\frac{1}{1+a_n}$$
$$\Rightarrow a_n^2+a_n-1=0$$
$$\Rightarrow a_n=\frac{\sqrt{5}-1}{2}$$

 

[lfdahl]

Spoiler

Substitute $a_n$ repeatedly i.e
$$a_n=\frac{1}{1+a_{n-1}}=\frac{1}{1+\frac{1}{1+a_{n-2}}}=\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}}$$
Since $n\rightarrow \infty$, we can write:
$$a_n=\frac{1}{1+a_n}$$
$$\Rightarrow a_n^2+a_n-1=0$$
$$\Rightarrow a_n=\frac{\sqrt{5}-1}{2}$$

Correct, short and consistent solution Pranav. Well done!(Nod)