sum of ((-4)^(n-1))/7^n from n = 1 to infinity
From wolfram|alpha I can conclude the solution is (1/11):
(http://www.wolframalpha.com/input/?i=sum…
However, I don't know how to get there? Here is what I did:
Sum( -4^(n-1) / 7^n =
Sum( (-4^n * -4^-1) / 7^n =
(-1/4)*Sum( -4^n / 7^n ) =
(-1/4)*Sum( (-4/7)^n ) =
(-1/4)*Sum( (-4/7)^n ) =
(1/4)*Sum( (4/7)^n )
We plug in n = 1 to get numerator value: -4^(1-1) / 7^1 = 1/7 =>
(1/7)/( 1 - (4/7) ) = 1/3
1/3 does not equal 1/11? Where did I go wrong? Thanks.
From wolfram|alpha I can conclude the solution is (1/11):
(http://www.wolframalpha.com/input/?i=sum…
However, I don't know how to get there? Here is what I did:
Sum( -4^(n-1) / 7^n =
Sum( (-4^n * -4^-1) / 7^n =
(-1/4)*Sum( -4^n / 7^n ) =
(-1/4)*Sum( (-4/7)^n ) =
(-1/4)*Sum( (-4/7)^n ) =
(1/4)*Sum( (4/7)^n )
We plug in n = 1 to get numerator value: -4^(1-1) / 7^1 = 1/7 =>
(1/7)/( 1 - (4/7) ) = 1/3
1/3 does not equal 1/11? Where did I go wrong? Thanks.
-
Is it (-4)^(n-1) or is it -4^(n - 1)? If it is truly the former, then you subtracted when you should have added.
(1/7)/(1 - (-4/7)) = (1/7)/(1 + 4/7) = 1/(7 + 4) = 1/11.
(1/7)/(1 - (-4/7)) = (1/7)/(1 + 4/7) = 1/(7 + 4) = 1/11.