Re: 審歛
※ 引述《min102257 (暱稱)》之銘言：
: ∞ 1
: Σ --------------- conv.?
: N=3 ln(lnN)
: (lnN)
: 有點複雜不知怎麼算
: 化成積分好像也積不出來
: 那應該是用 比較檢驗法??
: 麻煩高手了~ 感謝
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帶入 然後整個級數乘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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