does 3^(-2^n) converge linearly, superlinearly, or quadratically? why?

Answer 1

First, we rewrite this as 9^(-n), using a^(xy) = (a^x)^y.

The convergence of x_n to its limit x is of order p if |x_n+1-x| <= c|x_n-x|^p, for some constant c.

In this case, x=0, so we want to find p such that

|9^-(n+1)| <= c|9^(-n)|^p.

Since |9^-(n+1)| = (1/9)|9^(-n)| for all n, we conclude that c=(1/9) and p=1, so the convergence is linear.