• 올인원 · 1117418 · 04/16 23:35 · MS 2021

    lim(n→∞) a[n] = lim(n→∞) n²/(16n² - 4) = 1/16

    ∑(n=1~∞) (a[n] - 1/16)
    = ∑(n=1~∞) (a[n] - n²/(16n² - 4)) + ∑(n=1~∞) {n²/(16n² - 4) - 1/16}
    = 3/8 - ∑(n=1~∞) {n²/(16n² - 4) - 1/16}

    ∑(k=1~n) n²/(16n² - 4)
    = ∑(k=1~n) {1/16 + 1/4 * 1/(16n² - 4)}
    = ∑(k=1~n) {1/16 + 1/16 * 1/(2n + 1)(2n - 1)}
    = ∑(k=1~n) [1/16 + 1/32 * {1/(2n - 1) - 1/(2n + 1)}]
    = 1/16n + 1/32(1 - 1/(2n + 1))
    → ∑(k=1~n) {n²/(16n² - 4) - 1/16} = 1/32(1 - 1/(2n + 1))
    ∑(n=1~∞) {n²/(16n² - 4) - 1/16} = 1/32
    ∑(n=1~∞) (a[n] - 1/16) = 3/8 + 1/32 = 13/32

    lim(n→∞) {a[n] + ∑(k=1~n)(a[k] - 1/16)} = 1/16 + 13/32 = 15/32