Question

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)]

Answer 1

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

Answer 2

Other hint: this is a telescoping series.