推 math1209::-) 03/25 21:59
※ 引述《hanabiz (叛逆之臣也)》之銘言:
: show that 1 - 1/2 + 1/3 - 1/4 + ......... = log2
: thx
: 好像有兩種解法?
提供大一微積分課本習題的作法:
令S_n= 1 - 1/2 + 1/3 -1/4 +.......+ {(-1)^(n-1)}/n
H_n= 1 + 1/2 + 1/3 +1/4 +.......+ 1/n
T_n= 1 + 1/2 + 1/3 +1/4 +.......+ 1/n -ln(n)
(a)Show that S_2n = H_2n - H_n
H_2n - H_n = 1 + 1/2 + 1/3 +1/4 +...+ 1/2n - (1 + 1/2 + 1/3 +1/4 +...+ 1/n)
= 1 + 1/2 + 1/3 +1/4 +...+ 1/2n - 2(1/2 + 1/4 + 1/6 +1/8 +...+ 1/2n)
= 1 - 1/2 + 1/3 -1/4 +....+ {(-1)^(2n-1)}/2n
= S_2n
(b)Show that T_n has a limit (The value is called Euler's costant,γ)
因為f(x)=1/x 在x>0時遞減
畫圖可觀察到
n
∫1/x dx < 1 + 1/2 + 1/3 +...+ 1/n
1
=>ln(n) - 0 < 1 + 1/2 + 1/3 +...+ 1/n
=>1 + 1/2 + 1/3 +...+ 1/n - ln(n) = T_n 恆正
又 T_n - T_(n+1) = {ln(n+1)-ln(n)} - 1/(n+1)
可視為 {ln(n+1)-ln(n)} 的面積減去高為 1/(n+1) 底為1的長方形面積
從圖形看明顯 > 0
得 T_n - T_(n+1)>0
又 lim T_n = 0
n→∞
由交錯級數審斂可知級數T_n收斂 此極限稱為Euler's costant 以γ表示
(c)
From (b),we have H_n-ln(n) → γ as n→∞
and therefore H_2n-ln(2n) → γ as n→∞
lim [{H_2n-ln(2n)} - {H_n-ln(n)}] = lim {H_2n - H_n - (ln2 +ln(n)) + ln(n)}
n→∞ n→∞
= lim(S_n - ln2)
n→∞
= γ - γ = 0
由 lim(S_n - ln2) = 0
n→∞
即得 lim S_n = ln2 證畢.
n→∞
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◆ From: 218.166.57.7
※ 編輯: keith291 來自: 218.166.57.7 (03/25 21:52)