Re: [微分] 一題級數之斂散性判斷
∞ (lnn)^n ∞
Σ ───── = Σ a_n
n=1 n! n=1
Root:
lnn
(a_n)^(1/n) = ──────
(n!)^(1/n)
如果你有做過下列這個題目:
(n!)^(1/n) 1
lim ───── = ── (解法:取ln後發現是lnx在[0,1]的黎曼合)
n→∞ n e
n lnn
則 (a_n)^(1/n) = ───── ────
(n!)^(1/n) n
↓ ↓
e 0
Ratio:
a_(n+1) ln(n+1) (ln(n+1))^n
──── = ───── ──────
a_n n+1 (lnn)^n
↓ ↓
0 1 (why?)
reason for why:
(ln(n+1))^n
Let b_n = ──────
(lnn)^n
ln(n+1)
then ln( ─── ) → 0 (裡面趨近於1 by L')
ln(n+1) lnn
ln(b_n) = n ln ( ─── ) = ─────
lnn 1
── → 0
n
it's of the form 0/0
Using L'hospital rule , we have
1 1
lnn * ─── - ln(n+1) * ───
lnn n+1 n
─── * [ ──────────────── ]
ln(n+1) (lnn)^2
────────────────────────
-1
──
n^2
整理一下得:
lnn - n^2 1 1
─── * ──── * [ lnn * ─── - ln(n+1) * ─── ]
ln(n+1) (lnn)^2 n+1 n
↓ ↓ ↓
1 0 0-0 = 0
所以 ln(b_n) → 0
所以b_n → 1
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◆ From: 1.169.132.45
※ 編輯: znmkhxrw 來自: 1.169.132.45 (03/12 00:36)
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