Why do ratios of these Fibonacci-type sequences approach $\pi$?

Question

Define $A_n$ by $A_1=12$, $A_2=18$, and $A_n=A_{n-1}+A_{n-2}$ for $n\ge3$. Similarly define $B_n$ by $B_1=5$, $B_2=5$, and $B_n=B_{n-1}+B_{n-2}$ for $n\ge3$.

Terms of $A_n$: $12, 18, 30, 48, 78,\dots$

Terms of $B_n$: $5, 5, 10, 15, 25,\dots$

I found that dividing element $A_n$ by $B_n$ where $n$ approaches $\infty$ appears to result in:

$$\lim\limits_{n\to \infty}\left ( \frac{A_n}{B_n} \right ) = \pi$$

My question is, why does the ratio appear to converge towards $\pi$, and what is the significance of $5, 12, 18$ as to why this happens?

As pointed out by Michael, the correct limit is $$L =\lim\limits_{n\to \infty}\left ( \frac{A_n}{B_n} \right ) = \frac{6\phi^2}{5} = 3.14164078\dots$$ No wonder it looked familiar -- this approximation for $\pi$ was already found by *Ramanujan* and expressed as, $$L = \sqrt{\frac{9}{5}}+\frac{9}{5} = 3.14164078\dots$$ See this related post. — Tito Piezas III, Mar 02 '16 at 00:27

Empy2 · Accepted Answer · 2016-03-01T03:34:11.380

9

The correct ratio is $\frac{6\phi^2}5=3.14164079$, which is remarkably close to $\pi$. The $6/5$ comes because the starting values are $6,6$ in one and $5,5$ in the other. The $\phi^2$ comes because the sequences are shifted two steps from each other - both 6's appear before the 5's.

edited Mar 01 '16 at 03:34

answered Feb 29 '16 at 23:46

Empy2

50,853

1

"remarkably close" is a coincidence? (Or as close to a coincidence as math allows anything to be?) – fleablood Mar 01 '16 at 00:00
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I don't agree with your numerical result. The convergence is to $3.1416407865$. – Mar 01 '16 at 00:22
@YvesDaoust is right.... $6\phi^2 / 5 \approx 3.14164079.$ Still remarkably close to $\pi$. – mjqxxxx Mar 01 '16 at 00:34
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@YvesDaoust, it looks like Michael rounded $\phi$ to $1.618$ for the opening calculation. – Barry Cipra Mar 01 '16 at 00:43
Yes, that was my mistake. Fixed now. – Empy2 Mar 01 '16 at 03:33
@mjqxxxx: yep, remarkably close to $3.14$. – Mar 01 '16 at 07:51

score 4 · Answer 2 · answered Mar 01 '16 at 00:18

4

Recurrences of the form

$$T_n=T_{n-1}+T_{n-2}$$ are linear and known to have a general solution of the form $$T_n=C_0z_0^n+C_1z_1^n,$$ where $z_0,z_1$ are the roots of the "characteristic equation", $z^2=z+1$.

By the usual formulas,

$$z_0,z_1=\frac{1\pm\sqrt5}2.$$ Using the initial conditions,

$$T_0=C_0+C_1,\\T_1=C_0z_0+C_1z_1.$$

As $|z_0|>|z_1|$, the first term quickly dominates and

$$T_n\approx\frac{T_1-z_1T_0}{z_0-z_1}z_0^n.$$

In your case, for large $n$,

$$\frac{A_n}{B_n}\approx\frac{A_1-z_1A_0}{B_1-z_1B_0}=\frac{3\sqrt5+9}5=3.1416407864999\cdots$$

answered Mar 01 '16 at 00:18

+1 This close approximation was already known to Ramanujan and expressed by him as, $$\frac{3\sqrt{5}+9}{5} = \sqrt{\frac{9}{5}}+\frac{9}{5}$$ Just another example of his mathematical aesthetics. Kindly see this post. – Tito Piezas III Mar 02 '16 at 00:36
@TitoPiezasIII: it turns out that the solution of $\sqrt x+x=\pi$ is $1.7999649350783\cdots$. The next rational approximation of this number isn't so nice: $10267/5704$, giving only $8$ correct decimals. – Mar 02 '16 at 08:46

Why do ratios of these Fibonacci-type sequences approach $\pi$?

2 Answers2