# Limit of a specific sequence involving Fibonacci numbers.

Let, $\left\{F_n\right\}_{n=1}^\infty$ be the Fibonacci sequence, i.e, $F_1=1, F_2=1~\&~ F_{n+2}=F_{n+1}+F_n~\forall ~n \in \mathbb{Z}_+$

Let, $P_1=0, P_2=1$. Divide the line segment $\overline{P_n P_{n+1}}$ in the ratio $F_n:F_{n+1}$ to get $P_{n+2}$.

So, $P_{n+2}=\dfrac{F_n P_{n+1}+F_{n+1}P_n}{F_n+F_{n+1}}=\dfrac{F_n}{F_{n+2}}P_{n+1}+\dfrac{F_{n+1}}{F_{n+2}}P_n$

What is the limit of the sequence $\left\{P_n \right\}_{n=1}^\infty$ ?

$\textbf{Few things:}$ If we define,

\begin{eqnarray*} I_n &=& \left[P_n,P_{n+1}\right] \mathrm{,~if~} n \mathrm{~is ~odd~}\\ &=& [P_{n+1},P_n] \mathrm{,~if~} n \mathrm{~is ~even~} \end{eqnarray*} then we see $I_n \supseteq I_{n+1}~\forall~n \in \mathbb{Z}_+$ and $\lim \limits_{n \to \infty} |I_n|=0$

So, by Cantor's nested interval theorem, $\bigcap \limits_{n=1}^\infty I_n$ is singleton. Hence, $\lim \limits_{n \to \infty} P_n$ exists.

I tried a little bit, but I couldn't find the limit.

• Mathematica thinks the limit is 0.710855351429328416887694490384… Dec 4, 2015 at 22:28
• I'm not talking about numerical solution, isn't there a exact formula? @PatrickStevens Dec 4, 2015 at 22:29
• Isn't it clear that I don't know the answer? :P I'm trying to provide easy information that might help you or someone else to answer the question. Dec 4, 2015 at 22:31
• Ha ha :P . And thanks for the information :-) @PatrickStevens Dec 4, 2015 at 22:31
• Yeah. But, $P_1 \le P_3 \le \cdots \le P_{2n-1} \le \cdots\le \lim \limits P_n \le \cdots \le P_{2n} \le \cdots \le P_4 \le P_2$ Dec 5, 2015 at 1:46

The first few values of $P_n$ for $n \ge 2$ seem to be alternating sums of reciprocal Fibonacci numbers, starting with the second Fibonacci number (the second 1 in the sequence, so denominators go 1,2,3,5,8,13, etc.) $$P_2=\frac{1}{1},\ P_3=\frac{1}{1} - \frac12, \ P_4=\frac11-\frac12+\frac13, \\ P_5=\frac11-\frac12+\frac13-\frac15,\ P_6=\frac11-\frac12+\frac13-\frac15+\frac18$$ So the limiting value of $P_n$ would be the value of this alternating series. One would need to check that defining the $P_n$ this way makes them satisfy the recurrence in the posted question. I may try to work on that part. But it seems so much of a coincidence that it "has to" be true!
Anyway I did use the above method to go for some large $n$ values and got intervals which closed in on the numerical value found by Patrick Stevens in his comment.
• Still working on verifying the alternating reciprocal function satisfies the recurrence. I think by peeling off the last few terms and factoring out, it may come down to some known Fibonacci identity like $F_nF_{n+2}-F_{n+1}^2=\pm 1$ but I'm still lost in details. [If anyone cares to supply that (inductive) proof I'd appreciate (and cite) it.] Dec 5, 2015 at 2:52
• I have verified it. If we have, $P_n=\dfrac{1}{F_2}-\dfrac{1}{F_3}+\dfrac{1}{F_4}-\cdots +(-1)^n \dfrac{1}{F_n}$ and similar expression for $P_{n+1}$, from here we can find the expression for $F_{n+2}P_{n+2}=(F_n+F_{n+1})P_{n+2}$, from which we can see $P_{n+2}$ have a similar expression as $P_n$. Hence by induction we're done. @coffeemath your observation about $P_2, P_3, P_4, P_5, P_6$ was great. Dec 5, 2015 at 3:00
• wolfram says both $\sum \limits_{n=1}^\infty \dfrac{1}{F_{2n}}$ and $\sum \limits_{n=1}^\infty \dfrac{1}{F_{2n-1}}$ have closed forms, so the limit of $P_n$ have a closed form too. Reciprocal Fibonacci Constant link Dec 5, 2015 at 3:09