## anonymous 4 years ago find an expression for the nth partial sum Sn of the series summation of n=2 to infinity of 2[1/(2^n) -1/(2^(n+1)]

1. campbell_st

get a common denominator $2[(2^(n+1) - 2^n)/(2^n)(2^(n+1)]$ rewrite as $2[2^n(2 -1)/2^n(2^(n+1)]$ cancel the common factor 2^n gives $2/2^(n+1) = 2/(2\times2^n)$ cancel another common factor 2 gives 1/2^n

2. barrycarter

Other hint: this is a telescoping series.