ALG-B1-M05-P024 Decrease of a Radical Sequence

Set \(A=\sqrt[3^{n+1}]{a}\), \(B=\sqrt[3^{n+1}]{b}\), and factor \(B^3-A^3\).

Then \(B>A>1\), \(y_{n+1}=3^{n+1}(B-A)\), and \(y_n=3^n(B^3-A^3)=3^{n+1}(B-A)\frac{B^2+AB+A^2}{3}\).

Since \(A,B>1\), the average \(\frac{B^2+AB+A^2}{3}>1\). Therefore \(y_n>y_{n+1}\).

Rewrites the method of comparing consecutive terms through a difference of powers.