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NN321
Group Title
how do you find the sum from n=1 to infinity of n(1/4)^(n1)
 one year ago
 one year ago
NN321 Group Title
how do you find the sum from n=1 to infinity of n(1/4)^(n1)
 one year ago
 one year ago

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perl Group TitleBest ResponseYou've already chosen the best response.1
did you try using the ratio test
 one year ago

NN321 Group TitleBest ResponseYou've already chosen the best response.0
I thought the ratio test tells you if it is convergent or divergent not what the actual sum is. By the ratio test it is convergent.
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
dw:1353738999087:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
is that the question>
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
the expression inside looks close to the derivative of 1/4 ^n1
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
the answer is 16/9
 one year ago

NN321 Group TitleBest ResponseYou've already chosen the best response.0
how did you get that?
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
lets look at the partial sums
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
take the sum from n=1 to m
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
do you have a justification
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
dw:1353747498242:dw
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
dw:1353747595344:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
what about the limits ?
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
i could start the second limit at n = 2
 one year ago

NN321 Group TitleBest ResponseYou've already chosen the best response.0
how did you get 1/(1x)^2 from the step above?
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
dw:1353747789535:dw
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
take the derivative both sides
 one year ago

NN321 Group TitleBest ResponseYou've already chosen the best response.0
ok. cool. thank you
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
you can also take integral sometimes..
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
that formula is incorrect
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
dw:1353740583496:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
dw:1353740607290:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
very sloppy indeed
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
http://mathworld.wolfram.com/GeometricSeries.html
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
I did not notice your series start from 1, answer will be (16/9)1
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
dw:1353748518528:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
the answer is correct , it is 16/9
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
dw:1353741295190:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
the first term is zero, so we can ignore it
 one year ago

NN321 Group TitleBest ResponseYou've already chosen the best response.0
the original solution was correct. the left hand side is defined at n=1, the RHS doesn't have to be defined at 1 because its the formula for the sum of geometric series i.e. r=x.
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
dw:1353741624657:dw
 one year ago
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