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NN321

  • 3 years ago

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

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  1. perl
    • 3 years ago
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    did you try using the ratio test

  2. NN321
    • 3 years ago
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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.

  3. perl
    • 3 years ago
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    |dw:1353738999087:dw|

  4. perl
    • 3 years ago
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    is that the question>

  5. NN321
    • 3 years ago
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    yes

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

  7. perl
    • 3 years ago
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    the answer is 16/9

  8. NN321
    • 3 years ago
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    how did you get that?

  9. perl
    • 3 years ago
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    lets look at the partial sums

  10. perl
    • 3 years ago
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    take the sum from n=1 to m

  11. cinar
    • 3 years ago
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    16/9

  12. perl
    • 3 years ago
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    do you have a justification

  13. cinar
    • 3 years ago
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    |dw:1353747498242:dw|

  14. cinar
    • 3 years ago
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    derivative

  15. cinar
    • 3 years ago
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    x=1/4

  16. cinar
    • 3 years ago
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    |dw:1353747595344:dw|

  17. perl
    • 3 years ago
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    what about the limits ?

  18. cinar
    • 3 years ago
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    not needed

  19. perl
    • 3 years ago
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    i could start the second limit at n = 2

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

  21. cinar
    • 3 years ago
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    |dw:1353747789535:dw|

  22. cinar
    • 3 years ago
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    take the derivative both sides

  23. NN321
    • 3 years ago
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    ok. cool. thank you

  24. cinar
    • 3 years ago
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    (:

  25. cinar
    • 3 years ago
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    you can also take integral sometimes..

  26. perl
    • 3 years ago
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    that formula is incorrect

  27. perl
    • 3 years ago
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    |dw:1353740583496:dw|

  28. perl
    • 3 years ago
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    |dw:1353740607290:dw|

  29. perl
    • 3 years ago
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    very sloppy indeed

  30. cinar
    • 3 years ago
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    http://mathworld.wolfram.com/GeometricSeries.html

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

  32. cinar
    • 3 years ago
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    |dw:1353748518528:dw|

  33. perl
    • 3 years ago
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    the answer is correct , it is 16/9

  34. perl
    • 3 years ago
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    |dw:1353741295190:dw|

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

  36. NN321
    • 3 years ago
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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.

  37. perl
    • 3 years ago
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    |dw:1353741624657:dw|

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