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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.

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  • $\begingroup$ Mathematica thinks the limit is 0.710855351429328416887694490384… $\endgroup$ – Patrick Stevens Dec 4 '15 at 22:28
  • $\begingroup$ I'm not talking about numerical solution, isn't there a exact formula? @PatrickStevens $\endgroup$ – Mahbub Alam Dec 4 '15 at 22:29
  • $\begingroup$ 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. $\endgroup$ – Patrick Stevens Dec 4 '15 at 22:31
  • $\begingroup$ Ha ha :P . And thanks for the information :-) @PatrickStevens $\endgroup$ – Mahbub Alam Dec 4 '15 at 22:31
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    $\begingroup$ Yeah. But, $P_1 \le P_3 \le \cdots \le P_{2n-1} \le \cdots\le \lim \limits P_n \le \cdots \le P_{2n} \le \cdots \le P_4 \le P_2$ $\endgroup$ – Mahbub Alam Dec 5 '15 at 1:46
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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.

Now if the signs are dropped the terms all become positive and that constant has been discussed for example here and at Wolfram on the same topic it is said that the value of the sum for even indexed Fibonacci numbers is a known closed form constant. At the Wiki site the sum of all positive reciprocals is given a name but no known closed form for it seems to exist. There's a lot of material about it, though, like it is irrational as I recall. With the info about even indexed Fibonacci reciprocals it should at least be possible to express the alternating sum with the constant which is the sum of the positive reciprocals, and that would mean no hope for a closed form for the alternating sum either.

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  • $\begingroup$ That was a nice answer. Thanks $\endgroup$ – Mahbub Alam Dec 5 '15 at 2:20
  • $\begingroup$ 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.] $\endgroup$ – coffeemath Dec 5 '15 at 2:52
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    $\begingroup$ 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. $\endgroup$ – Mahbub Alam Dec 5 '15 at 3:00
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    $\begingroup$ 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 $\endgroup$ – Mahbub Alam Dec 5 '15 at 3:09
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    $\begingroup$ @Mahbub The odd indexed one's closed form has theta functions in it, so in that sense isn't such a closed form as the even index sum which is all in terms of the golden ratio and powers of it. $\endgroup$ – coffeemath Dec 5 '15 at 3:21

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