A sequence is defined as $a_{n+1}=1+\frac{1}{a_n}$ for $n \ge 1$ with $a_{1}=1$

Question

Will it converge and to what limit?

I found that the hint given in the book is that interpret each term $a_n=\frac{F_{n+1}}{F_n}$ where $F_{n}$ is the $n$ th Fibonacci number.

But I would be highly obliged if anyone poses a general approach towards this problem.

Also, I can show that $a_n$ is bounded above by $2$ and below by $1$..So, for the limit to exist all we need to show is that the sequence is monotonic, but I am not sure how to show this.

Suppose you know the limit $a$ exists, then it has to hold $$ a= 1 + \frac1a$$. — Stefan Hante, May 17 '16 at 10:49
(For $a\neq 0$), Also, the sequence is not monotonic starting from $a_2$. — Stefan Hante, May 17 '16 at 10:55

score 1 · Answer 1 · answered May 17 '16 at 11:07

If it has a limit $\alpha$, then we have $\alpha=1+\frac{1}{\alpha}$, so $\alpha^2-\alpha-1=0$ and hence $\alpha=\frac{\sqrt5\pm1}{2}$. But we clearly have $a_n>1$ for all $n$, so the limit (if it exists) must be $\alpha=\frac{\sqrt5+1}{2}$. Note that $\frac{1}{\alpha}=\alpha-1$.

We have $a_{n+1}-\alpha=1-\alpha+\frac{1}{a_n}=\frac{1}{a_n}-\frac{1}{\alpha}=\frac{\alpha-a_n}{\alpha a_n}$. Note that $a_n>1$, so $|a_{n+1}-\alpha|<\frac{1}{\alpha}|a_n-\alpha|$. Since $\alpha>1$ this implies that $\lim_{n\to\infty}|a_n-\alpha|=0$. Hence the sequence tends to $\alpha$.

score 0 · Answer 2 · answered May 17 '16 at 10:57

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To simplify this problem consider two different series and prove they are converging , odd terms and even . Then prove they converge to same answer . Represent $a_{n+2}=f(a_{n})$

answered May 17 '16 at 10:57

avz2611

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A sequence is defined as $a_{n+1}=1+\frac{1}{a_n}$ for $n \ge 1$ with $a_{1}=1$

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