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NN321
 2 years ago
how do you find the sum from n=1 to infinity of n(1/4)^(n1)
NN321
 2 years ago
how do you find the sum from n=1 to infinity of n(1/4)^(n1)

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perl
 2 years ago
Best ResponseYou've already chosen the best response.1did you try using the ratio test

NN321
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.

perl
 2 years ago
Best ResponseYou've already chosen the best response.1the expression inside looks close to the derivative of 1/4 ^n1

perl
 2 years ago
Best ResponseYou've already chosen the best response.1lets look at the partial sums

perl
 2 years ago
Best ResponseYou've already chosen the best response.1take the sum from n=1 to m

perl
 2 years ago
Best ResponseYou've already chosen the best response.1do you have a justification

perl
 2 years ago
Best ResponseYou've already chosen the best response.1i could start the second limit at n = 2

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

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0take the derivative both sides

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0you can also take integral sometimes..

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

perl
 2 years ago
Best ResponseYou've already chosen the best response.1the answer is correct , it is 16/9

perl
 2 years ago
Best ResponseYou've already chosen the best response.1the first term is zero, so we can ignore it

NN321
 2 years ago
Best ResponseYou've already chosen the best response.0the 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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