Round reciprocals of Fibonacci numbers

Published:

The Fibonacci numbers $F_n$ are roughly equal to $\phi^n / \sqrt{5}$, where $\phi = (1 + \sqrt{5}) / 2$ is the golden ratio. In fact, they’re so close that rounding $\phi^n / \sqrt{5}$ will give you the right answer:

\begin{align*} \frac{\phi^5}{\sqrt{5}} &= 4.9596747752\dots \\ F_5 &= 5. \end{align*}

This implies that the series $\sum_{k = 1}^\infty \frac{1}{F_k}$ converges because it is essentially a geometric series with ratio $\phi^{-1}$. In particular, this means that the tails of this series must go to zero, which means that the reciprocals of the tails get big. More explicitly, the sequence

$S(n) = \left(\sum_{k = n}^\infty \frac{1}{F_k}\right)^{-1}$

goes to infinity. Here are the first few terms:

n S(n)
2 0.4237
3 0.7354
4 1.1629
5 1.8991
6 3.0623
7 4.9615
8 8.0238

Do you notice anything interesting about this table? Let’s see it one more time, with a new column.

n S(n) round(S(n))
2 0.42404 0
3 0.73537 1
4 1.1629 1
5 1.8991 2
6 3.0623 3
7 4.9615 5
8 8.0238 8

If you round the tails, you get the Fibonacci numbers again! That is, it looks like

$\mathrm{round}\left(\left(\sum_{k = n}^\infty \frac{1}{F_k}\right)^{-1}\right) = F_{n - 2}.$

This turns out to be correct. In fact, it turns out to be correct for a pretty broad class of sequences. Here, I’m going to show a quick proof, summarize what we know about these sequences, and point out where there are gaps in our knowledge.

Why does this work?

The intuition here is that, since $F_n$ is basically $\phi^n / \sqrt{5}$, something like this should be true:

$\sum_{k \geq n} \frac{1}{F_k} \approx \sum_{k \geq n} \frac{\sqrt{5}}{\phi^k} = \frac{\sqrt{5}}{\phi^{n - 1} (\phi - 1)}.$

Then we reciprocate both sides, and get something useful to play with. The proof just makes this intuition more rigorous. It uses the asymptotic identity $(1 + O(a(n)))^{-1} = 1 + O(a(n))$, valid whenever $a(n) \to 0$.

Proof. By Binet’s formula, $F_n = A \phi^n + B \psi^n$ for some constants $A$ and $B$, where $\phi$ and $\psi$ are the golden ratio and its conjugate, respectively. Therefore,

\begin{align*} \frac{1}{F_k} &= (A \phi^k)^{-1} (1 + O((\psi / \phi)^k))^{-1} \\ &= (A \phi^k)^{-1} (1 + O((\psi / \phi)^k)) \\ &= (A \phi^k)^{-1} + O((\psi / \phi^2)^k). \end{align*}

If we sum this over $k \geq n$, then we obtain

$\sum_{k \geq n} \frac{1}{F_k} = (A \phi^{n - 1} (\phi - 1))^{-1} + O((\psi / \phi^2)^n).$

Reciprocating yields

\begin{align*} \left(\sum_{k \geq n} \frac{1}{F_k}\right)^{-1} &= ((A \phi^{n - 1} (\phi - 1))^{-1} + O((\psi / \phi^2)^n))^{-1} \\ &= A \phi^{n - 1} (\phi - 1) (1 + O((\psi / \phi)^n)) \\ &= A \phi^{n - 1} (\phi - 1) + O(\psi^n). \end{align*}

Now, note that $A \phi^{n - 1} (\phi - 1) = F_n - F_{n - 1} + O(\psi^n)$, and that $F_n - F_{n - 1} = F_{n - 2}$. This gives us

$\left(\sum_{k \geq n} \frac{1}{F_k}\right)^{-1} = F_{n - 2} + O(\psi^n).$

Since $|\psi| < 1$, eventually the left-hand side is within $1/2$ of $F_{n - 2}$, so rounding the left-hand side for sufficiently large $n$ gives $F_{n - 2}$. $\blacksquare$

The key insights were something like this:

1. The Fibonacci numbers have characteristic equation $(x - \phi)(x - \psi)$.
2. One root, $\phi$, is greater than $1$. The other root, $\psi$, has absolute value less than $1$.
3. The reciprocal sums are $F_{n - 2} + O(\psi^n)$.

It turns out that this proof applies to nearly any sequence whose characteristic polynomial satisfies these properties. The theorem goes something like this.

Theorem. Let $c(n)$ be an integer-valued sequence which satisfies a linear recurrence with constant coefficients. If the characteristic equation of this recurrence has one root outside the unit circle and all other roots strictly inside the unit circle, then

$\mathrm{round} \left( \left( \sum_{k \geq n} \frac{1}{c(k)} \right)^{-1} \right) = c(n) - c(n - 1)$

for sufficiently large $n$.

What don’t we know?

Fix an integer-valued sequence $c(n)$ which satisfies a linear recurrence with characteristic polynomial $P(x)$. Let

$S(n) = \left( \sum_{k \geq n} \frac{1}{c(k)} \right)^{-1}.$

The question is what happens to $\mathrm{round}(S(n))$.

1. If the largest root of $P(x)$ is outside the unit circle and the smaller roots are inside the unit circle, then $S(n)$ rounds to $c(n) - c(n - 1)$ for sufficiently large $n$. This is the previous theorem I mentioned.

There are lots of papers which say this, then provide an example of its application. They mostly have the flavor of, “look at an infinite family of polynomials I found which satisfies this property that we need.”

2. When does $S(n)$ not round to $c(n) - c(n - 1)$? Is the previous condition about the roots necessary? What if there are roots on the unit circle?

3. When does $\mathrm{round}(S(n))$ satisfy a linear recurrence with constant coefficients? Perhaps it does not equal $c(n) - c(n - 1)$, but it might still satisfy a recurrence of some kind. This would be a neat closure property of C-finite sequences.

These last two questions seem open.

Roots on the unit circle

Let me demonstrate that “one root is outside the unit circle and the others are inside” is not a necessary condition. Take the example sequence $c(n) = 2^n - 1$, which has the characteristic polynomial $(x - 1)(x - 2)$. Let

$S(n) = \left( \sum_{k \geq n} \frac{1}{2^k - 1} \right)^{-1}.$

There is very convincing data that $S(n)$ will round to $2^{n - 1}$:

n S(n)
2 0.622
3 1.648
4 3.658
5 7.662
6 15.665
7 31.666

However, the previous proof does not work. If we try to follow the argument again, we’ll get

$S(n) = 2^{n - 1} + O(1),$

which is not good enough to say what $S(n)$ will round to. This is because one root of the polynomial is on the unit circle rather than inside it.

Nevertheless, there is an argument. We just need to be more careful with our asymptotics. Start by writing

\begin{align*} \frac{1}{2^k - 1} &= \frac{1}{2^k} \left(1 + \frac{2^{-k}}{1 - 2^{-k}}\right) \\ &= \frac{1}{2^k} + \frac{2^{-2k}}{1 - 2^{-k}}. \end{align*}

If we sum this over $k \geq n$, then we get

$S(n)^{-1} = 2^{-n + 1} + \sum_{k \geq n} \frac{2^{-2k}}{1 - 2^{-k}}.$

Call that last sum $E(n)$. In the previous proof, we essentially say $E(n) = O(2^{-2n})$ and call it a day, which ends up giving us an $O(1)$ error. By being more careful, we can determine the constant on $2^{-2n}$ and thereby improve our final error term. It looks like this:

\begin{align*} \sum_{k \geq n} \frac{2^{-2k}}{1 - 2^{-k}} &= \sum_{k \geq n} (2^{-2k} + O(2^{-3k})) \\ &= \frac{2^{-2n}}{1 - 2^{-2}} + O(2^{-3n}). \end{align*}

So, in fact, $E(n) = \frac{4}{3} 2^{-2n} + O(2^{-3n})$. If we stick with $E(n)$, we have

\begin{align*} S(n) &= (2^{-n + 1} + E(n))^{-1} \\ &= 2^{n - 1} (1 + 2^{n - 1} E(n))^{-1} \\ &= 2^{n - 1} (1 - 2^{n - 1} E(n) + O(2^{2n} E(n)^2)). \end{align*}

Now we can plug-in our improved estimate for $E(n)$:

\begin{align*} S(n) &= 2^{n - 1} - 2^{2n - 2} (2^{-2n} (1 - 2^{-2})^{-1} + O(2^{-3n})) + O(2^{-n}) \\ &= 2^{n - 1} - \frac{1}{3} + O(2^{-n}). \end{align*}

The sequence $S(n)$ is almost exactly a third away from $2^{n - 1}$, so eventually we’ll have $\mathrm{round}(S(n)) = 2^{n - 1}$.

The takeaway is that sometimes $S(n)$ will round to $c(n) - c(n - 1)$ even when the characteristic equation of $c(n)$ has a root on the unit circle. So we still don’t know a necessary condition for this rounding behavior.