jmc

Question

Let x be a real number, x > 1. Compute _n = 0^ 1x^2^n - x^-2^n.

Accepted Answer

We can write 1x^2^n - x^-2^n = x^2^nx^2^n + 1 - 1.Let y = x^2^n. Then align* x^2^nx^2^n + 1 - 1 &= yy^2 - 1 &= (y + 1) - 1y^2 - 1 &= y + 1y^2 - 1 - 1y^2 - 1 &= 1y - 1 - 1y^2 - 1 &= 1x^2^n - 1 - 1x^2^n + 1 - 1. align*Thus, the sum telescopes: _n = 0^ 1x^2^n - x^-2^n = ( 1x - 1 - 1x^2 - 1 ) + ( 1x^2 - 1 - 1x^4 - 1 ) + ( 1x^4 - 1 - 1x^8 - 1 ) + = 1x - 1.