How do you find the fourth partial sum of #Sigma 8(-1/4)^n# from n=1 to #oo#?
1 Answer
Jul 7, 2017
Explanation:
Let use denote
# S_n = sum_(r=1)^n \ 8(-1/4)^r#
So then. the fourth partial sum, is
# S_4 = sum_(r=1)^4 \ 8(-1/4)^r#
# " " = 8(-1/4)^1 + 8(-1/4)^2 + 8(-1/4)^3 + 8(-1/4)^4#
# " " = 8(-1/4) + 8(1/16) + 8(-1/64) + 8(1/256)#
# " " = 8{-1/4 + 1/16-1/64+1/256)#
# " " = 8{-51/256)#
# " " = -51/32#
