The series $$\sum_{k=0}^{\infty} \dfrac1{F_{2^k}} = \dfrac{7-\sqrt5}2$$where $F_m$ denotes the $m^{th}$ Fibonacci number is called the Millin series. In this post, we will look at a generalization of this result for recurrences of the form
$$a_{n+2} = b a_{n+1} + a_n$$
where $a_0=0$, $a_1 = 1$ and $b \in \mathbb{R}$. Note that in the case of the Fibonacci numbers, we have $b = 1$.
Claim: For any $b \in \mathbb{R}$, we have $$\sum_{k=0}^{\infty} \dfrac1{a_{2^k}} = 1 + \dfrac2b + \dfrac{b - \sqrt{b^2+4}}2 \,\,\,\,\, (\heartsuit)$$
Note that if we prove the above claim, then setting $b=1$ gives us
$$\sum_{k=0}^{\infty} \dfrac1{F_{2^k}} = 1 + 2 + \dfrac{1-\sqrt5}2 = 3 + \dfrac{1-\sqrt5}2 = \dfrac{7-\sqrt5}2$$which is the Millin series.
Proof: Before we prove this, let us first obtain a result for the finite sum counterpart in $(\heartsuit)$.
----------------------------------------------------------------------------------------------------------------------------------------------------------
Lemma: If we have a sequence given by the recurrence,
$$a_{n+2} = ba_{n+1} + a_n,$$ with $a_0 =0 $ and $a_1 = 1$, we then have
$$\boxed{\color{blue}{\displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^N-1}}{a_{2^N}}}}$$
Proof: Let us write out a few terms of this sequence, we get
$$a_0 = 0, a_1 = 1, a_2 = b, a_3 = b^2 + 1, a_4 = b^3 + 2b, \cdots$$
The proof is by induction on $N$. For $N=1$, we have the left hand side to be $$\dfrac1{a_1} + \dfrac1{a_2} = 1 + \dfrac1b$$ while the right hand side is $$1 + \dfrac2b - \dfrac{a_1}{a_2} = 1 + \dfrac2b - \dfrac1{b} = 1 + \dfrac1b$$
For $N=2$, we have the left hand side to be $$\dfrac1{a_1} + \dfrac1{a_2} + \dfrac1{a_4} = 1 + \dfrac1b + \dfrac1{b^3 + 2b}$$ while the right hand side is $$1 + \dfrac2b - \dfrac{a_3}{a_4} = 1 + \dfrac2b - \dfrac{b^2+1}{b^3+2b} = 1 + \dfrac1b + \dfrac1b - \dfrac{b^2+1}{b^3+2b} = 1 + \dfrac1b + \dfrac1{b^3+2b}$$ Hence, it holds for $N=1$ and $N=2$. Now lets go ahead with induction now. Assume the result is true for $N=m$, i.e., we have
$$\sum_{k=0}^{m} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^m-1}}{a_{2^m}}$$
Now $$\sum_{k=0}^{m+1} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^m-1}}{a_{2^m}} + \dfrac1{a_{2^{m+1}}}$$ Hence, we want to show that
$$ - \dfrac{a_{2^m-1}}{a_{2^m}} + \dfrac1{a_{2^{m+1}}} = -\dfrac{a_{2^{m+1}-1}}{a_{2^{m+1}}}, \,\, \text{i.e.}, \,\, \dfrac1{a_{2^{m+1}}} + \dfrac{a_{2^{m+1}-1}}{a_{2^{m+1}}} = \dfrac{a_{2^m-1}}{a_{2^m}}, \,\, \text{i.e.}, \,\, a_{2^m}(1+a_{2^{m+1}-1}) = a_{2^m-1} a_{2^{m+1}} \,\,\,\, (\star)$$
which can be verified using the recurrence. In fact $(\dagger)$, a slightly more general version of $(\star)$, which is easier to check is true.
$$a_{2k}(1+a_{4k-1}) = a_{2k-1} a_{4k} \,\,\,\, (\dagger),\,\, \text{i.e.}, \,\, a_{2k-1} a_{4k} - a_{2k} a_{4k-1} = a_{2k} \,\,\,\, (\dagger)$$
Hence, we get that
$$\boxed{\color{red}{\displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^N-1}}{a_{2^N}}}}$$
----------------------------------------------------------------------------------------------------------------------------------------------------------
Now letting $N \to \infty$, we get what we want immediately. This is so since from the recurrence we get that
$$\dfrac{a_{n+2}}{a_{n+1}} = b + \dfrac{a_n}{a_{n+1}}$$
If we have $\displaystyle \lim_{n \to \infty} \dfrac{a_n}{a_{n+1}} = L$, then we get that
$$\dfrac1L = b + L$$ and since $L>0$, we have $L = \dfrac{\sqrt{b^2+4}-b}2$. Hence,
$$\boxed{\color{red}{\displaystyle \sum_{k=0}^{\infty} \dfrac1{a_{2^k}} = \lim_{N \to \infty} \displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \lim_{N \to \infty} \dfrac{a_{2^N-1}}{a_{2^N}} = 1 + \dfrac2b - L = 1 + \dfrac2b + \dfrac{b-\sqrt{b^2+4}}2}}$$
$$a_{n+2} = b a_{n+1} + a_n$$
where $a_0=0$, $a_1 = 1$ and $b \in \mathbb{R}$. Note that in the case of the Fibonacci numbers, we have $b = 1$.
Claim: For any $b \in \mathbb{R}$, we have $$\sum_{k=0}^{\infty} \dfrac1{a_{2^k}} = 1 + \dfrac2b + \dfrac{b - \sqrt{b^2+4}}2 \,\,\,\,\, (\heartsuit)$$
Note that if we prove the above claim, then setting $b=1$ gives us
$$\sum_{k=0}^{\infty} \dfrac1{F_{2^k}} = 1 + 2 + \dfrac{1-\sqrt5}2 = 3 + \dfrac{1-\sqrt5}2 = \dfrac{7-\sqrt5}2$$which is the Millin series.
Proof: Before we prove this, let us first obtain a result for the finite sum counterpart in $(\heartsuit)$.
----------------------------------------------------------------------------------------------------------------------------------------------------------
Lemma: If we have a sequence given by the recurrence,
$$a_{n+2} = ba_{n+1} + a_n,$$ with $a_0 =0 $ and $a_1 = 1$, we then have
$$\boxed{\color{blue}{\displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^N-1}}{a_{2^N}}}}$$
Proof: Let us write out a few terms of this sequence, we get
$$a_0 = 0, a_1 = 1, a_2 = b, a_3 = b^2 + 1, a_4 = b^3 + 2b, \cdots$$
The proof is by induction on $N$. For $N=1$, we have the left hand side to be $$\dfrac1{a_1} + \dfrac1{a_2} = 1 + \dfrac1b$$ while the right hand side is $$1 + \dfrac2b - \dfrac{a_1}{a_2} = 1 + \dfrac2b - \dfrac1{b} = 1 + \dfrac1b$$
For $N=2$, we have the left hand side to be $$\dfrac1{a_1} + \dfrac1{a_2} + \dfrac1{a_4} = 1 + \dfrac1b + \dfrac1{b^3 + 2b}$$ while the right hand side is $$1 + \dfrac2b - \dfrac{a_3}{a_4} = 1 + \dfrac2b - \dfrac{b^2+1}{b^3+2b} = 1 + \dfrac1b + \dfrac1b - \dfrac{b^2+1}{b^3+2b} = 1 + \dfrac1b + \dfrac1{b^3+2b}$$ Hence, it holds for $N=1$ and $N=2$. Now lets go ahead with induction now. Assume the result is true for $N=m$, i.e., we have
$$\sum_{k=0}^{m} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^m-1}}{a_{2^m}}$$
Now $$\sum_{k=0}^{m+1} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^m-1}}{a_{2^m}} + \dfrac1{a_{2^{m+1}}}$$ Hence, we want to show that
$$ - \dfrac{a_{2^m-1}}{a_{2^m}} + \dfrac1{a_{2^{m+1}}} = -\dfrac{a_{2^{m+1}-1}}{a_{2^{m+1}}}, \,\, \text{i.e.}, \,\, \dfrac1{a_{2^{m+1}}} + \dfrac{a_{2^{m+1}-1}}{a_{2^{m+1}}} = \dfrac{a_{2^m-1}}{a_{2^m}}, \,\, \text{i.e.}, \,\, a_{2^m}(1+a_{2^{m+1}-1}) = a_{2^m-1} a_{2^{m+1}} \,\,\,\, (\star)$$
which can be verified using the recurrence. In fact $(\dagger)$, a slightly more general version of $(\star)$, which is easier to check is true.
$$a_{2k}(1+a_{4k-1}) = a_{2k-1} a_{4k} \,\,\,\, (\dagger),\,\, \text{i.e.}, \,\, a_{2k-1} a_{4k} - a_{2k} a_{4k-1} = a_{2k} \,\,\,\, (\dagger)$$
Hence, we get that
$$\boxed{\color{red}{\displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^N-1}}{a_{2^N}}}}$$
----------------------------------------------------------------------------------------------------------------------------------------------------------
Now letting $N \to \infty$, we get what we want immediately. This is so since from the recurrence we get that
$$\dfrac{a_{n+2}}{a_{n+1}} = b + \dfrac{a_n}{a_{n+1}}$$
If we have $\displaystyle \lim_{n \to \infty} \dfrac{a_n}{a_{n+1}} = L$, then we get that
$$\dfrac1L = b + L$$ and since $L>0$, we have $L = \dfrac{\sqrt{b^2+4}-b}2$. Hence,
$$\boxed{\color{red}{\displaystyle \sum_{k=0}^{\infty} \dfrac1{a_{2^k}} = \lim_{N \to \infty} \displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \lim_{N \to \infty} \dfrac{a_{2^N-1}}{a_{2^N}} = 1 + \dfrac2b - L = 1 + \dfrac2b + \dfrac{b-\sqrt{b^2+4}}2}}$$
