Rational or irrational?

Let $F(0)=0, \; F(1)=\frac{3}{2}$ and
$$F(n)=\frac{5}{2}F(n-1)-F(n-2), \;\; n \geq 2$$

Examine if the number $\displaystyle \sum_{n=0}^{\infty} \frac{1}{F(2^n)}$ is rational or not.

IMC 2015/ 1st Round/ 3rd problem

Solution



The characteristic equation of the recurrance relation is simply $x^2=\frac{5}{2}x- 1$ and we note that the roots are $x_1=\frac{1}{2}, \; x_2=2$. Hence by taking into account the initial values we get that:

$$F(n)=2^n -\frac{1}{2^n}$$

We can now see that the sum telescopes. Hence:

$$\sum_{n=0}^{\infty}\frac{1}{2^{2^n}- \frac{1}{2^{2^n}}}= \sum_{n=0}^{\infty}\frac{2^{2^n}}{2^{2^{n+1}}-1}= \sum_{n=0}^{\infty}\frac{2^{2^n}+1-1}{2^{2^{n+1}}-1}= \\
=\sum_{n=0}^{\infty}\left [ \frac{1}{2^{2^n}-1}- \frac{1}{2^{2^{n+1}}-1} \right ]= \frac{1}{2-1}=1$$

and the exercise comes to an end.