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The evaluation,

$$\sum_{n=0}^\infty \frac{1}{F_{2^n}}=\frac{7-\sqrt{5}}{2}=\left(\frac{1-\sqrt{5}}{2}\right)^3+\left(\frac{1+\sqrt{5}}{2}\right)^2$$

was recently asked in a post by Chris here.

I like generalizations, and it turns out this is not a unique feature of the Fibonacci numbers. If we use the Pell numbers $P_m = 1,2,5,12,29,70,\dots$ then the sum is also an algebraic number of deg 2. In general, it seems for any positive rational b, then,

$$\sum_{n=0}^\infty \frac{1}{\frac{1}{\sqrt{b^2+4}}\left( \left(\frac{b+\sqrt{b^2+4}}{2}\right)^{2^n}-\left(\frac{b-\sqrt{b^2+4}}{2}\right)^{2^n}\right)}=1+\frac{2}{b}+\frac{b-\sqrt{b^2+4}}{2}$$

where Fibonacci numbers are just the case b = 1, the Pell numbers b = 2, and so on. (For negative rational b, then one just uses the positive case of $\pm\sqrt{b^2+4}$.)

Anyone knows how to prove/disprove the conjectured evaluation?

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The induction approach mentioned in the linked post ought to work; the key point is that you can use the index-doubling formula of the associated sequence - which comes from the matrix representation of the sequence's shift operator - to explicitly express the partial sums. – Steven Stadnicki Jun 13 '12 at 15:40
(also, is there a typo in your second formula? You use the expressions $\frac{1}{2}(b^2\pm\sqrt{b^2+4})$ where I assume you mean $\frac{1}{2}(b\pm\sqrt{b^2+4})$...) – Steven Stadnicki Jun 13 '12 at 15:42
Oops,you are right. I've corrected the typo. – Tito Piezas III Jun 13 '12 at 15:45
up vote 17 down vote accepted

Your conjecture is indeed right. Before proving your conjecture, let us obtain an intermediate result first. Let us prove the following claim first.


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 $$ - \dfrac{a_{2^m-1}}{a_{2^m}} + \dfrac1{a_{2^{m+1}}} = -\dfrac{a_{2^{m+1}-1}}{a_{2^{m+1}}}$$ i.e. $$\dfrac1{a_{2^{m+1}}} + \dfrac{a_{2^{m+1}-1}}{a_{2^{m+1}}} = \dfrac{a_{2^m-1}}{a_{2^m}}$$ 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. $$a_{2k}(1+a_{4k-1}) = a_{2k-1} a_{4k} \,\,\,\, (\dagger)$$ 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 see that your conjecture is indeed right. 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, $$\boxed{\color{red}{\displaystyle \sum_{k=0}^{\infty} \dfrac1{a_{2^k}} = \lim_{N \to \infty} \displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \lim_{N \to \infty} \dfrac{a_{2^N-1}}{a_{2^N}} = 1 + \dfrac2b - L = 1 + \dfrac2b + \dfrac{b}2 -\dfrac{\sqrt{b^2+4}}2}}$$


After some googling, I found out that a similar result is true for a more general class of recurrences of the form $$a_{n+1} = P a_n + Q a_{n-1}$$ See this article for more details.

Also, try googling Millin series for more details.

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+1 for going through the slog of the math and especially for the information about further generalizations - great stuff! – Steven Stadnicki Jan 21 '13 at 18:02
@StevenStadnicki Yes. The trick was to get the right claim for finite $N$ i.e. to get $$\color{blue}{\displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^N-1}}{a_{2^N}}}.$$ Once we had it, the rest is just plain simple induction and then to let $N \to \infty$. – user17762 Jan 21 '13 at 18:05
Thanks, Marvis! (I am on vacation, so didn't see your answer till now.) – Tito Piezas III Jan 25 '13 at 3:10
And also for the links. They were most helpful. – Tito Piezas III Jan 25 '13 at 3:39

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