[passer-by]

สังเกตว่า $ a_n = \frac{1}{2 \sqrt 3} \left( x^n - x^{-n} \right) =\frac{1}{2 \sqrt 3} \left( \frac{x^{2n}-1}{x^n} \right)$ เมื่อ $ x= 2+\sqrt{3} $

ดังนั้น $$ \begin{array}{rcl} \sum_{n=0}^\infty \frac{1}{ a_{ 2^n} } & = & 2\sqrt{3}\left( \frac{x}{x^2-1}+\frac{x^2}{x^4-1}+\frac{x^4}{x^8-1}+\cdots \right ) \\ &=& 2\sqrt{3}\left( \frac{x+1-1}{x^2-1}+\frac{x^2+1-1}{x^4-1}+\frac{x^4+1-1}{x^8-1}+\cdots \right ) \\
&=& 2\sqrt{3}\left(( \frac{1}{x-1}-\frac{1}{x^2-1})+(\frac{1}{x^2-1}-\frac{1}{x^4-1})+( \frac{1}{x^4-1}-\frac{1}{x^8-1})+\cdots \right ) \\&=& 2\sqrt{3} \left (\frac{1}{x-1} \right ) \\&=& 2\sqrt{3} \left (\frac{1}{\sqrt{3}+1} \right ) \\ &=& 3- \sqrt{3} \end{array} $$