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Infinite series

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data

    a) consider the infinite series (k=1) sum (inf) [(k+1)^(1/2) - (k)^(1/2)]
    expand and simplify the nth partial sum. determine wether the oartial sums S_n converge as n-> inf

    b) determine all the numbers x in R so that the infinite series

    (k=0) sum (inf) [x^(k)/(k!)]


    c) determine wheter the series

    (k=1) sum (inf) [k/(k^3 + 1)] converges or diverges.

    2. Relevant equations

    3. The attempt at a solution

    a) I wrote out the terms of the nth partial sums

    S_1 = (2)^(1/2) - 1
    S_2 = (2)^(1/2) - 1 + (3)^(1/2) - (2)^(1/2) = (3)^(1/2) - 1
    S_3 = (2)^(1/2) - 1 + (3)^(1/2) - (2)^(1/2) + (4)^(1/2) - (3)^(1/2) = (4)^(1/2) - 1

    therefore, the nth partial sum simplifies down to

    S_n = (n+1)^(1/2) - 1

    and converges to infinity as n-> inf

    b) (k=0) sum (inf) [x^(k)/(k!)]

    looking for all x in R so it converges

    I used the ratio test to get

    | [(x)^(k+1)/(k+1)!] / [x^(k)/(k!)] | < 1

    then I get -(k+1) < x < (k+1)

    so if x is between those values, the series converges.

    c) Converges by the comparison test

    Hi, can someone let me know if I got these right? thanks!
  2. jcsd
  3. Nov 15, 2008 #2


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    Hi squaremeplz! :smile:

    (have a square-root: √ and a sigma: ∑ and an infinity: ∞ :wink:)

    a) and c) are ok.
    But k goes up to ∞ :redface:

    (and anyway you should be familiar with this series :wink:)
  4. Nov 16, 2008 #3
    ah, so it converges for all values of x?
  5. Nov 16, 2008 #4


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    Yup! :biggrin:

    And its value is …
    ? :wink:
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