Write out the first few terms of the geometric series, Summation from n equals 0 to infinity (StartFraction x minus 4 Over 4 EndFraction )Superscript n , to find a and r, and find the sum of the series. Then, express the inequality, StartAbsoluteValue r EndAbsoluteValue less than 1, in terms of x and find the values of x for which the inequality holds and the series converges.
Accepted Solution
A:
Answer:a=1 converges for 0<x<8Step-by-step explanation:Given that a geometric series infinite is given in summation form as[tex]\Sigma _0^\infty (\frac{x-4}{4} )^n[/tex]Here I term is when n =0So a= I term = [tex](\frac{x-4}{4} )^0=1[/tex]r = common ratio = [tex]\frac{x-4}{4}[/tex]This series converges only if [tex]|r|<1[/tex][tex]|\frac{x-4}{4}|<1\\-4<x-4<4\\0<x<8[/tex]For 0<x<8, we get absolute value of r is less than 1.Hence series converges for [tex]0<x<8[/tex]