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Math Help - convergence series

  1. #1
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    convergence series

    Hey All,

    I have two questions on series which I have done and just need someone to check them for me. Thank you

    1) find range of value of x for the convergence series:
    x + 2^4.x^2/2! + 3^4.x^3/3! + 4^4.x^4/4! + ...

    so:
    Un = n^4x^n/n!

    Un + 1 = (n+1)^4x^{n+1}/(n+1)!

    (Un + 1)/(Un) = (n+1)^4x^{n+1}/(n+1)! . n!/n^4x^n

    (x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4 therefore:
    (n+1)^4 = n^4 + 4n^3 +6n^2 + 4n + 1

    (Un + 1)/(Un) = (n^4 + 4n^3 +6n^2 + 4n + 1)x^n.x^1/n! + 1! . n!/n^4x^n

    Simplifying:
    (Un + 1)/(Un) = (4n^3 +6n^2 + 4n + 1)x^1/1!

    Lim n→∞ (Un + 1)/(Un) = (4 + 6/n + 4/n^2 + 1/n^3)x/n^3 / 1!/n^3

    = (4 + 0 + 0 + 0)0 / 0 = 0

    converges

    2) investigate convergence of series
    1/1.2 + 1/2.2^2 + 1/3.2^3 + 1/4.2^4 + ...

    so:
    Un = 1 / n2^n
    Un + 1 = 1 / (n+1)2^{n+1}

    (Un + 1)/(Un) = 1 / (n+1)2^n.2^1 . n2^n/1

    (Un + 1)/(Un) = n2^n / (n+1)2^n.2^1

    (Un + 1)/(Un) = n / (n+1)2

    Lim n→∞ (Un + 1)/(Un) = 0 / (0 +1)2 = 0

    converges
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  2. #2
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    Hello, dadon!

    Your algebra is a bit off . . . and your limits are way off.



    1) Find the inverval of convergence for:

    . . x + (2^
    4x^2)/2! + (3^4x^3)/3! + (4^4x^4)/4! + . . .
    Un+1 . . . . (n+1)^4x^{n+1} . . .n! . . . . . .(n+1)^4 .x^{n+1} . .n!
    ------ . = . ------------------------------ . = . ---------------------------
    . U
    n . . . . . . . . (n+1)! . . . .n^4x^n . . . . . n^4 . . . .x^n . .(n+1)!


    . . . . .(n+1)^
    4 . .x . . . . .(n+1) . . . . . n + 3n + 3n + 1
    . . = . --------------- . = . --------x . = . ----------------------x
    . . . . . . n^
    4 . . n+1 . . . . . n^4 . . . . . . . . . . n^4


    . . . . . . . . . . . . . . . . . . . . . . .1/n + 3/n + 3/n + 1/n^
    4
    Divide top and bottom by n^
    4: . -------------------------------x
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1


    Then: .lim (1/n + 3/n + 3/n + 1/n^
    4)x . = . 0
    . . . . n→∞

    Therefore, the series converges for all values of x.



    2) investigate convergence of the series:
    . . 1/(12) + 1/(22^2) + 1/(32^3) + 1/(42^4) + . . .

    U
    n+1 . . . . . . . . 1 . . - . . .n2^n . . - . . n - . 1
    ------ . = . ------------------------- . = . ---------
    . U
    n . . . . .(n+1)2^{n+1} . . 1 . . . . . .n + 1 .2

    . . . . . . . . . . . . . . . . . . . . . . . 1 . . . .1
    Divide top and bottom by n: . --------- --
    . . . . . . . . . . . . . . . . . . . . . 1 + 1/n . 2

    . . . . . . . . . . .1 . - - .1 . . . . 1
    Then: .lim .---------- -- . = . --
    . . . . x→∞ .1 + 1/n . .2 . . . . 2


    Therefore, the series converges.

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