推 PONYINNCCU :犀利 04/21 21:42
※ 引述《bajifox (嘖)》之銘言：
: n (-1)^k+1 n
: Σ ------- ( ) = A
: k=1 k k
: 帶了一些數字進去依然看不出為何如此
: 謝謝
f(x) = Σ[k;1,n] (1/k) C(n,k) x^k
f(0) = 0
then A = -f(-1)
f'(x) =Σ[k,1,n] C(n,k) x^{k-1}
x f'(x) = (1+x)^n - 1
f'(x) = (1+x)^n / x - 1/x
let 1+x = u
f'(u-1) = -u^n / (1-u) + 1/(1-u)
= (1-u^n)/(1-u)
= 1+u+u^2+...+u^{n-1}
f(u-1) = u + u^2/2 + u^3/3 + ... + u^n / n + C
f(0) = 0 then C = -(1 + 1/2 + ... + 1/k)
A = -f(-1) = -C
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