We have a sequence of numbers defined recursively by $$u_{n+1}=1+\frac{1}{u_n},$$for $n\geqslant 1$. It is also given that $u_1=1$. Find the limit $l$ of the sequence $u_1,u_3,u_5,\dots$.

So I said, $u_1,u_3,u_5,\dots$ is given by $u_{2n+1}$for $n\geqslant 0$. Then, $$\lim_{n\to\infty}u_{2n+1}=1+\frac{1}{\lim_{n\to\infty}u_{2n}}.$$Now, at $n\to\infty$, $u_{2n}=u_{2u+1}=l$, and so $$l=1+1/l \Rightarrow l=\frac{1+\sqrt{5}}{2},$$since $l>0$.

Can somebody please explain to me what's wrong with my reasoning (because it really 'feels' wrong)?


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    $\begingroup$ I think the final answer is not wrong. However, why does $\lim_{n\rightarrow\infty} u_{2n}=\lim_{n\rightarrow\infty} u_{2n+1}$ ? $\endgroup$ – Amr Jun 19 '13 at 12:39
  • $\begingroup$ Also what if these limits don't exist in the first place $\endgroup$ – Amr Jun 19 '13 at 12:39
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    $\begingroup$ If you have established the existence of the limit, then your approach is fine. The only thing: you should express $u_{2n+1}$ in terms of $u_{2n-1}$, not $u_{2n}$ (as "odd" and "even" subsequences may, for example, not have the same limit). $\endgroup$ – Start wearing purple Jun 19 '13 at 12:40

Here's a more solid reasoning:

$$u_{2 n+1} = 1+\frac{1}{u_{2 n}} = 1+\frac{1}{1+\frac{1}{u_{2 n-1}}}$$

which means that

$$u_{2 n+1} = \frac{1+2 u_{2 n-1}}{1+u_{2 n-1}}$$

Now assuming the limit exists, apply the same reasoning as above and the same result emerges.

  • $\begingroup$ To prove the limit exists, you can use the above formula to show that $1\leq u_{2n+1} \leq \frac{1+\sqrt{5}}{2}$ for all $n\in\mathbb{N}$ (by recurrence, using the monotonicity of the function $f\colon x\mapsto \frac{1+2x}{1+x}$). Then, show that because of this $(u_{2n+1})_n$ is non-decreasing (by computing $u_{2n+1}-u_{2n-1}$). You'll then have, by monotone convergence, that the sequence $(u_{2n+1})_n$ converges. $\endgroup$ – Clement C. Jun 19 '13 at 13:27

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