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Where does this infinite series converge:

Sigma (k = 1 to infinity): 1/(9k^2 + 3k - 2)

  • Math -

    term(k) = 1/[(3k+2)(3k-1)]

    so S(1) = 1/(5 x 2) = 1/10
    S(2) = 1/10 + 1/(8 x 5) = 1/8 or 2/16
    S(3) = 1/8 + 1/88 = 12/88 = 3/22
    S(4) = 3/22 + 1/154 = 1/7 or 4/28

    it appears we have a nice pattern here and
    S(n) = n/(6n+4)

    This type of question usually comes up with the topic of "induction".

    You would now have to prove that this conjecture is true by induction

  • Math -

    Thanks Reiny! I couldn't factor that polynomial. Once there, the rest follows the book example

    Each term in the sequence: 1/(9k^2 + 3k - 2) = 1/((3k+2)(3k-1)) = 1/3*(1/(3k-1) - 1/(3k+2))
    When you take a sum of those terms, it's a telescoping series where the -1/(3k+2) cancels the next +1/(3k-1) term.
    The sum of a partial series from terms 1 to n = 1/3*(1/2 - 1/(3n + 2)). Limit to infinity = 1/6

  • Math -

    for your additional information:

    You might want to remember that in these kind of telescoping series, the two factors are of this pattern:
    (mk + a)(mk + b) where a - b = m

    in our case they were (3k+2)(3k-1)
    notice 2-(-1) = 3

  • Math -

    Also notice that if you simplify your
    1/3*(1/2 - 1/(3n + 2)) you get my n/(6n+4)

    and Limit n/(6n+4) as n ---> ∞ = 1/6

  • Math -

    jko[j

  • Math -

    5_13+7_61=11_20

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