For any t in R, by partial sum of geometric series,
1 - t + t^2 - t^3 + ... + (-1)^n*(t^n) = (1 - (-1)^(n+1) * t^(n+1) ) / (1+t)
Thus
(1+t)^-1 = 1 - t + t^2 - t^3 + ... + (-1)^n*(t^n) + (-1)^(n+1)*t^(n+1)/(1+t)
Integrating from 0 to x,
log(1+x) = x - x^2/2 + x^3/3 - ... + (-1)^n*x^(n+1)/(n+1) + R_n(x)
where R_n(x) = int_0^x {(-1)^(n+1)*t^(n+1)/(1+t)} dt.
Consider when 0 <= x <= 1,
|R_n(x)| <= int_0^x {t^(n+1) / (1+t)} dt
<= int_0^x t^(n+1) dt
= x^(n+2) / (n+2)
tends to 0 as n goes to infinity.
Hence the series converges to log(1+x) also when x = 1.
※ 引述《jacky840816 (說好的女朋友呢?)》之銘言：
: 請問為什麼ln(1+x)的定義範圍是x小於等於1。
: 如果等於1，不就不能帶入1/1+x=1-x+x^2……這個公式了嗎？
: 無窮等比公式的定義不是公比要小於1嗎？
: 謝解惑
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 219.79.202.191
※ 文章網址: http://www.ptt.cc/bbs/Math/M.1398184065.A.5B2.html