# Sum of infinite series and polynomial equation

Let $$a=\sum_{n=1}^\infty(1 \mod \phi^{-n})2^{n/3},$$ where $\phi = \frac{1 + \sqrt{5}}{2}$ is the golden ratio and $1 \mod x=1-x\lfloor x^{-1}\rfloor$.

How can I prove that $a$ satisfies the equation $$2255 a^6 - 2340 a^5 - 3174 a^4 - 672 a^3 + 180 a^2 + 36 a - 36=0?$$

-
You could plus in the series and work out the terms, then prove that the result converges to 0, but this would be a lot of work. I do not however see a more elegant approach. –  Alex Becker Jan 17 '12 at 2:13
Do you have some reason to think that $a$ satisfies that equation? The more you tell us about what's behind the problem, the easier you make it for someone to help you with it. –  Gerry Myerson Jan 17 '12 at 3:42
@GerryMyerson The only reason was that numerical approximation suggested so (I used a computer program to guess the coefficients of the polynomial) –  Vladimir Reshetnikov Jan 21 '12 at 22:17

The following approach should work, although the details are messy. Let $\psi = (1 - \sqrt{5})/2 = -1/\phi$ be the Galois conjugate of $\phi$. Then the quantity $b_n = \phi^n + \psi^{n}$ is an integer, because it's both an algebraic integer and invariant under the Galois automorphism $\sqrt{5} \mapsto -\sqrt{5}$. But $|\psi|<1$, and $\psi$ is negative, so that $\lfloor{\phi^n}\rfloor$ equals $b_n$ if $n$ is odd, or $b_n - 1$ if $n$ is even. Hence we have $$(1\,\mathrm{ mod }\,\phi^{-n}) = 1 - \frac{\lfloor{\phi^n}\rfloor}{\phi^n} = 1 - \frac{\phi^n + \psi^n - e_n}{\phi^n} = \frac{-\psi^n+ e_n}{\phi^n}$$ where $e_n$ is 0 if $n$ is odd, 1 if $n$ is even. Now that we've gotten rid of the floor, we can substitute this expression into the sum, which produces some geometric series which can easily be evaluated. The result is clearly an algebraic number in $\mathbb{Q}(\phi, 2^{1/3})$, and therefore has degree at most 6. Finding the exact minimal polynomial will be messy though.
Here are more details: Since $\psi^n = (-\phi)^{-n}$, the sum becomes $$\sum_{n=1}^{\infty} -(-2^{1/3} \phi^{-2})^n + (2^{2/3} \phi^{-2})^n$$ Both of the quantities in parentheses above have absolute value less than 1, so the two separate geometric series converge individually. Letting $u = -2^{1/3} \phi^{-2}$ and $v = 2^{2/3}\phi^{-2}$, the sum is equal to $$a = \frac{v}{1-v} - \frac{u}{1-u} \approx 1.86511275789174$$ which, at least numerically, appears to satisfy the polynomial given: $$2255 a^6 - 2340 a^5 - 3174 a^4 - 672 a^3 + 180 a^2 + 36 a - 36 \approx -4.8 \cdot 10^{-12}.$$ I also numerically calculated the Galois conjugates of $a$ and then used those values to compute the characteristic polynomial, which gives the above polynomial. (All calculations done in Sage.)
You don't need to compute the minimal polynomial; the question is to show that a polynomial of degree $6$ is satisfied, so it suffices to take the characteristic polynomial of our algebraic number acting by left multiplication on $\mathbb{Q}(\phi, \sqrt[3]{2})$, and this reduces in principle to computing a determinant. –  Qiaochu Yuan Jan 17 '12 at 4:16