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NN321

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

  • one year ago
  • one year ago

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  1. perl
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    did you try using the ratio test

    • one year ago
  2. 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.

    • one year ago
  3. perl
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    |dw:1353738999087:dw|

    • one year ago
  4. perl
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    is that the question>

    • one year ago
  5. NN321
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    yes

    • one year ago
  6. perl
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    the expression inside looks close to the derivative of 1/4 ^n-1

    • one year ago
  7. perl
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    the answer is 16/9

    • one year ago
  8. NN321
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    how did you get that?

    • one year ago
  9. perl
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    lets look at the partial sums

    • one year ago
  10. perl
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    take the sum from n=1 to m

    • one year ago
  11. cinar
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    16/9

    • one year ago
  12. perl
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    do you have a justification

    • one year ago
  13. cinar
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    |dw:1353747498242:dw|

    • one year ago
  14. cinar
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    derivative

    • one year ago
  15. cinar
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    x=1/4

    • one year ago
  16. cinar
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    |dw:1353747595344:dw|

    • one year ago
  17. perl
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    what about the limits ?

    • one year ago
  18. cinar
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    not needed

    • one year ago
  19. perl
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    i could start the second limit at n = 2

    • one year ago
  20. NN321
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    how did you get 1/(1-x)^2 from the step above?

    • one year ago
  21. cinar
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    |dw:1353747789535:dw|

    • one year ago
  22. cinar
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    take the derivative both sides

    • one year ago
  23. NN321
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    ok. cool. thank you

    • one year ago
  24. cinar
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    (:

    • one year ago
  25. cinar
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    you can also take integral sometimes..

    • one year ago
  26. perl
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    that formula is incorrect

    • one year ago
  27. perl
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    |dw:1353740583496:dw|

    • one year ago
  28. perl
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    |dw:1353740607290:dw|

    • one year ago
  29. perl
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    very sloppy indeed

    • one year ago
  30. cinar
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    http://mathworld.wolfram.com/GeometricSeries.html

    • one year ago
  31. cinar
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    I did not notice your series start from 1, answer will be (16/9)-1

    • one year ago
  32. cinar
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    |dw:1353748518528:dw|

    • one year ago
  33. perl
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    the answer is correct , it is 16/9

    • one year ago
  34. perl
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    |dw:1353741295190:dw|

    • one year ago
  35. perl
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    the first term is zero, so we can ignore it

    • one year ago
  36. 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.

    • one year ago
  37. perl
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    |dw:1353741624657:dw|

    • one year ago
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