※ 引述《min102257 (昵称)》之铭言： : ∞ 1 : Σ --------------- conv.? : N=3 ln(lnN) : (lnN) : 有点复杂不知怎麽算 : 化成积分好像也积不出来 : 那应该是用 比较检验法?? : 麻烦高手了~ 感谢 divergent. since ln(ln(n)) is increasing , ln(n) is increasing so ln(ln(n)) ln(n) is increasing to inf 1 so ─────── decreasing to 0 ln(ln(n)) ln(n) Cauchy condensation principle can be applied 2^n consider ───────── (n都用2^n带入 然後整个级数乘2^n) ln(nln(2)) nln(2) 2^n => ───────────── (ln(n) + ln(ln(2))) nln(2) 2^n => ────────────────────────── denoted by b_n ln(n) ln(ln(2)) ln(n) ln(ln(2)) n * n * ln(2) * ln(2) ~~~~~~~~~~ ↓ ln(ln(2)) n by root test, (b_n)^(1/n) 2 = ──────────────────────────────── (ln(n))/n (ln(ln(2)))/n (ln(ln(2)))/n (ln(ln(2)))/n n * n * n * ln(2) ~~~~~~~~~~ ~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~ ↓ ↓ ↓ ↓ as n→inf, 1 1 1 1 so summation of b_n diverges imply the original question is divergent --

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