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I tried solving this recursive function using a linear combination method, but was unable to come to a conclusion. I do not know where to start finding the limit

Did
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Kevin McDonough
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    Hint: Compute the six first terms. What do you observe? – Did Sep 25 '15 at 21:07
  • General method, by the way: assuming the limit exists and is $x$, it satisfies $x = 1+\frac{1}{x}$ (by continuity). However, as Did says, the first thing you should always do with a sequence is to calculate the first few terms to see if you can spot a pattern. – Patrick Stevens Sep 25 '15 at 21:37
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    If you don't know how to use Did's hint:

    By computing the first terms of the sequence, one can note that $$x_n=\dfrac{f_{n+2}}{f_{n+1}},$$ where $f_i$ is the $i$-th term in the Fibonacci sequence and it is well known that $\lim\limits_{n\to\infty}\dfrac{f_{n+1}}{f_n}=\dfrac{1+\sqrt5}{2}.$

     – CIJ Sep 25 '15 at 22:11
  • FWIW, the idea in my hint is to realize that $(x_{2n})$ is increasing and $(x_{2n+1})$ is decreasing and $1\le x_{2k}<x_{2i+1}\le2$ for every $(k,i)$... and that these basic observations are awfully close to provide a full solution without any prior familiarity with Fibonacci sequences. – Did Sep 27 '15 at 07:21

1 Answers1

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Your ability to answer this quickly will depend on whether you're familiar with the Fibonacci numbers.

If you're not you can still use elementary methods to calculate the limit: Firstly however we need to show that the limit does indeed exist. To do this one could show its Cauchy. I did this by firstly showing by induction that: $ 3/2 \le x_n \le 2$ for all $n \ge 1$. (I'll omit this since it was just a standard inducion, but I can add it in if you want?)

I then used the definition of Cauchy with $n \ge m \ge N$ and calculated: \begin{align} |x_n - x_m| &= \left | 1 + \frac{1}{x_{n-1}} - x_m \right | \\ & = \left |1+\frac{1}{x_{n-1}} - 1 - \frac{1}{x_{m-1}} \right| \\ & = \frac{|x_{n-1} -x_{m-1}|}{x_nx_m} \end{align} Now we can iterate this procedure (assuming we are not in the trivial case of n=m) to obtain: \begin{align} |x_n - x_m| &= \frac{|x_{n-m} -x_{0}|}{\prod_{i=n-m+1}^n x_i \prod_{j=1}^m x_j}\\ & = \frac{|x_{n-m} - 1|}{\prod_{i=n-m+1}^n x_i \prod_{j=1}^m x_j}\\ & = \frac{1}{\prod_{i=n-m+1}^n x_i \prod_{j=1}^m x_j}* \frac{1}{x_{n-m-1}} \end{align}

Now using our bounds from before we deduce that: $$|x_n - x_m| \le \left(\frac{2}{3} \right)^{m}$$ (The $\left(\frac{2}{3} \right)^{m}$ simply comes from counting the number of elements on the bottom of the final fraction and tacking a crude bound)

Which allows us to deduce $x_n$ is Cauchy. By completeness of $\mathbb{R}$ we deduce $x_n$ is convergent.

Now using our bounds we know the limit is between $3/2$ and $2$ so there is no issue with $1/x_n$ 'blowing up' so we can take the limit of both sides giving (if $x_n \rightarrow x$) that $$ x = 1 + 1/x$$ You can now solve this like a normal quadratic to find $x$. Hope this helps!

Zestylemonzi
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