upgrayed 發問時間: 科學數學 · 10 年前

limit comparison test

use the limit comparison test to determine the convergence of following series

sigma(n=1 to inf) ( n + 4 ) / ( 2n^3 - n + 1 )

已更新項目:

theorem : Limit Comparison Test

suppose that sigma(n=1 to inf) a_n and sigma(n=1 to inf) b_n have positive terms

and that a_n/b_n -> L as n -> inf , where L ≠ 0

if sigma(n=1 to inf) b_n is convergent , then sigma(n=1 to inf) a_n is convergent

1 個解答

評分
  • linch
    Lv 7
    10 年前
    最佳解答

    Let an = (n+4) / (2n^3-n+1), bn = 1/n^2

    lim_(n to inf) an/bn

    = lim_(n to inf) n^2(n+4) / (2n^3-n+1)

    = lim_(n to inf) (n^3+4n^2) / (2n^3-n+1)

    = lim_(n to inf) (1 + 4/n) / (2 - 1/n^2 + 1/n^3)

    = 1/2

    and

    sigma(n=1 to inf) bn = sigma(n=1 to inf) 1/n^2 is a p-series, with p=2>1,

    hence sigma(n=1 to inf) bn converges.

    By the limit comparison test, sigma(n=1 to inf) (n+4) / (2n^3-n+1) converges.

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