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Let $\{a_n\}$ be a positive sequence which satisfies $a_{n+2}=a_{n+1}+a_n$ for $n=1,2,\ldots$. Let $z_n=a_{n+1}/a_n$. How can I prove that $\lim_{n\rightarrow\infty}z_n$ exists?

I looked at $z_{n+1}=1+1/z_n$, but I still don't know how to go from here.

Martin Sleziak
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PJ Miller
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2 Answers2

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Corrected version.

By hypothesis the $z_n$ are positive, so the recurrence

$$z_{n+1}=1+\frac1{z_n}$$

tells you that $z_n>1$ for $n\ge 2$. It follows that $1<z_n<2$ for $n\ge3$. If the sequence converges, it must converge to some $L\in[1,2]$. Moreover, that $L$ must be a fixed point of the function

$$f(x)=1+\frac1x\;:$$

$f$ is continuous, so it must satisfy

$$L=\lim_{n\to\infty}z_{n+1}=\lim_{n\to\infty}f(z_n)=f(L)\;.$$

Solving

$$L=1+\frac1L$$

for the unique solution in the interval $[1,2]$, we see that the only possible limit is

$$\varphi=\frac{1+\sqrt5}2\approx1.618\;.$$

Note that if $z>\varphi$, then $$f(z)=1+\frac1z<1+\frac1\varphi=\varphi\;,$$ while if $z<\varphi$, then $$f(z)=1+\frac1z>1+\frac1\varphi=\varphi\;.$$ That is, the sequence is alternately above and below $\varphi$.

Suppose that $1<a\le x<y$. The mean value theorem says that there is a $u\in[x,y]$ such that

$$\left|\frac{f(x)-f(y)}{x-y}\right|=|f\,'(u)|=\frac1{u^2}\le\frac1{a^2}\;.\tag{1}$$

You can easily check that if $1<z<2$, then $\frac32<f(z)<2$, so $z_n>\frac32$ for $n\ge 4$. We may therefore take $a=\frac32$ in $(1)$ to see that for each $n\ge 4$,

$$\left|\frac{z_{n+2}-z_{n+1}}{z_{n+1}-z_n}\right|\le\frac1{(3/2)^2}=\frac49$$

and hence

$$|z_{n+2}-z_{n+1}|\le\frac49|z_{n+1}-z_n|\;.$$

To finish, just combine this with the fact that the terms are alternatively larger and smaller than $\varphi$ to see that if $z_n<\varphi$, then

$$z_n<z_{n+2}<z_{n+4}<\ldots<\varphi<\ldots<z_{n+5}<z_{n+3}<z_{n+1}$$

with $\lim\limits_{k\to\infty}(z_{n+2k+1}-z_{n+2k})=0$: this shows that the sequence converges to $\varphi$.

Brian M. Scott
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  • I agree that $f$ is decreasing. How does it follow that $z_{n}$ is decreasing? – PJ Miller Jun 09 '13 at 04:40
  • @PJMiller: Sorry: that was a bad slip on my part. In fact the negative derivative of $f$ implies that the sequence alternates increasing and decreasing. I slightly more complicated argument is needed; it’ll take me a few minutes to fix it. – Brian M. Scott Jun 09 '13 at 05:00
  • Thanks. I don't think we need the fact that the terms alternate, do we? Having $|z_{n+2}-z_{n+1}|\le\frac49|z_{n+1}-z_n|$ should be enough to conclude that the sequence converges. – PJ Miller Jun 09 '13 at 14:04
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    @PJMiller: You can do it either way. The alternation tells you immediately that both the odd and the even subsequences converge, and the contraction then tells you that their limits must be the same. The other way you use the fact that a geometric series with ratio less than $1$ converges to conclude that the sequence is Cauchy and therefore converges. – Brian M. Scott Jun 09 '13 at 20:10
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HINT:

So, we have
$$\frac{a_{n+2}}{a_{n+1}}=1+\frac1{\frac{a_{n+1}}{a_n}}$$

$$\text{Now, }\lim_{n\to\infty}\frac{a_{n+2}}{a_{n+1}}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=y\text{(say)}$$

$$\implies y=1+\frac1y\text{ where } y>0 $$

lab bhattacharjee
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