how do you find the sum from n=1 to infinity of n(1/4)^(n-1)

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how do you find the sum from n=1 to infinity of n(1/4)^(n-1)

Mathematics
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did you try using the ratio test
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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Other answers:

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

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