NN321
how do you find the sum from n=1 to infinity of n(1/4)^(n-1)
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did you try using the ratio test
NN321
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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.
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|dw:1353738999087:dw|
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is that the question>
NN321
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yes
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the expression inside looks close to the derivative of 1/4 ^n-1
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the answer is 16/9
NN321
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how did you get that?
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lets look at the partial sums
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take the sum from n=1 to m
cinar
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16/9
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do you have a justification
cinar
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|dw:1353747498242:dw|
cinar
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derivative
cinar
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x=1/4
cinar
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|dw:1353747595344:dw|
perl
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what about the limits ?
cinar
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not needed
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i could start the second limit at n = 2
NN321
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how did you get 1/(1-x)^2 from the step above?
cinar
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|dw:1353747789535:dw|
cinar
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take the derivative both sides
NN321
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ok. cool. thank you
cinar
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(:
cinar
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you can also take integral sometimes..
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that formula is incorrect
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|dw:1353740583496:dw|
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|dw:1353740607290:dw|
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very sloppy indeed
cinar
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I did not notice your series start from 1, answer will be (16/9)-1
cinar
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|dw:1353748518528:dw|
perl
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the answer is correct , it is 16/9
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|dw:1353741295190:dw|
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the first term is zero, so we can ignore it
NN321
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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.
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|dw:1353741624657:dw|