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Scott
Scott asked in Science & MathematicsMathematics · 9 years ago

Calculus II help. Have a test coming?

Use the Ratio test to determine convergence:

infinite(sigma)k=1... k^4 2^-k

2 Answers

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  • The Integral ∫
    Lv 7
    9 years ago
    Favorite Answer

    using ratio test :

    ∞

    Σ [ k^4 / 2^(k) ]

    k = 1

    (a_(k + 1) / (a_k) )

    [ (k + 1)^4 / 2^(k + 1) ] ÷ [ k^4 / 2^(k) ]

    [ (k^4 + 4k^3 + 6k^2 + 4k + 1) / 2^(k) 2^1) ] * [ 2^(k) / k^4 ]

    [ (k^4 + 4k^3 + 6k^2 + 4k + 1) / 2) ] * [ 1 / k^4 ]

    [ (k^4 + 4k^3 + 6k^2 + 4k + 1) / ( 2 k^4 ) ]

    lim [ (k^4 + 4k^3 + 6k^2 + 4k + 1) / ( 2 k^4 ) ]

    n-->∞

    lim [ (k^4 + 4k^3 + 6k^2 + 4k + 1) / ( 2 k^4 ) ] <----- large in charge

    k-->∞

    lim (k^4) / ( 2 k^4 )

    k-->∞

    lim (1/2) = (1/2) < 1 ( convergent )

    k-->∞

    ========

    free to e-mail if have a question

  • Brenda
    Lv 7
    9 years ago

    k^4*2^(-k)

    lim ak+1 / ak

    (k+1)^4 *2(-k+1)] / [ k^4 *2^(-k)]

    after simplification:

    (k+1)^4 / 2k^4

    lim [ (k+1)^4 / 2k^4] as k→ ∞ is1/2.

    If L<1 then the series converges.

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