Re: [微積] 台大數研90高微考古題

[JohnMash (Paul)]

※ 引述《tasukuchiyan (Tasuku)》之銘言：
: 4.(b)
: Let a(n)>0, a(n+1)/a(n)<=(1-2/n) for n>=3.
: Show that the series of a(n) is convergent.



  consider

  lim_{x→∞} [ln x - ln(x-2)]/[ln (x+1) - ln x]

  = lim_{x→∞}  [1/x - 1/(x-2)] / [1/(x+1) - 1/x]

  = lim_{x→∞} (x-2-x)/[x(x-2)] / {(x-x-1)/[(x+1)x]}

  = 2

  hence, there exists N such that

  [ln n - ln (n-2)] > (1.5) [ ln (n+1) - ln n] whenever n > N

   n/(n-2) > [(n+1)/n]^(1.5)

   1-2/n = (n-2)/n < [n/(n+1)]^{1.5}

   that is,

   a(n+1)/a(n) < [n/(n+1)]^{1.5}

   hence,

   a(N+2) < a(N+1) [(N+1)/(N+2)]^{1.5}

   a(N+3) < a(N+2) [(N+2)/(N+3)]^{1.5} < a(N+1) [(N+1)/(N+3)]^1.5

   .....

   then

   a(N+2) + a(N+3) + ....

  < a(N+1) (N+1)^{1.5} [ 1/(N+2)^{1.5} + 1/(N+3)^{1.5} + 1/(N+4)^{1.5} + ...]

   Convergent,

   Done.



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推 tasukuchiyan:感謝，不過這要怎麼想到的? 多做自然就會嗎?
→ bineapple   :這題應該還有(a)小題吧 (a)應該就是提示
→ tasukuchiyan:我也這麼想，但是我想不到要怎麼用(a)去證明(b)
→ tasukuchiyan:(a)是說log(1-x)<=-x for 0<=x<1 and
→ tasukuchiyan:log[(1-1/2)(1-1/3)…(1-1/n)]<=-(1/2+1/3+…+1/n)
→ tasukuchiyan:這樣(a)和(b)有關聯嗎?
→ dogy007     :其實前面的 (a) 誤導我們用了較難的解法
→ dogy007     :很容易證出 a_n < 6a_3 /(n(n-1))
→ dogy007     :少打了等號，然後和 sum 1/(n(n-1)) 比較審斂
推 tasukuchiyan:喔喔~感謝，不過我怎麼算出a(n)<=2a(3)/((n-1)(n-2))
→ dogy007     :嗯，我好像不小心有弄錯了