✔ 最佳答案
π/4-(1/2)ln2
2009-06-20 14:13:47 補充:
For each positive integer k,let a_k=Σ(n=1~k)(-1)^(n-1)/[(2n)(2n-1)]
Σ(n=1~k)(-1)^(n-1)/[(2n)(2n-1)]
=Σ(n=1~k)(-1)^(n-1)[1/(2n-1)-1/(2n)]
=[Σ(n=1~k)(-1)^(n-1)/(2n-1)]-[Σ(n=1~k)(-1)^(n-1)/(2n)]-------(*)
Claim 1: Σ(n=1~∞)(-1)^(n-1)/n=ln2
Proof: For |x|<1,|t|<1,∫_[0,t] Σ(n=1~∞)(-1)^(n-1)*x^(n-1)dx
=∫_[0,t] Σ(n=1~∞)(-x)^(n-1)dx
=∫_[0,t] dx/(1+x)
=ln(1+t)
又Σ(n=1~∞)(-1)^(n-1)/n is converges,故由Abel's limit theorem
lim(t->1-)ln(1+t)=lim(t->1-)∫_[0,t] Σ(n=1~∞)(-1)^(n-1)*x^(n-1)dx
=lim(t->1-)Σ(-1)^(n-1)∫_[0,t]x^(n-1)dx
=lim(t->1-)Σ(-1)^(n-1)*t^n/n
=Σ(-1)^(n-1)/n
Claim 2 Σ(-1)^(n-1)/(2n-1)=π/4
proof :Similarly, for |x|<1,|t|<1
∫_[0,t]Σ(-1)^(n-1)*x^(2n-2)dx
=∫_[0,t]Σ(-x^2)^(n-1)dx
=∫_[0,t]1/(1+x^2)dx
=tan^(-1)(t)
又Σ(-1)^(n-1)/(2n-1) is converges, again by Abel's limit Theorem
π/4=lim(t->1-)tan^(-1)(t)=lim(t->1-)∫_[0,t]Σ(-1)^(n-1)x^(2n-2)dx
=lim(t->1-)Σ(-1)^(n-1)∫_[0,t] x^(2n-2)dx
=lim(t->1-)Σ(-1)^(n-1)t^(2n-1)/(2n-1)
=Σ(-1)^(n-1)/2n-1
故原式=lim(k->∞)a_k=Σ(-1)^(n-1)/(2n-1)-(1/2)Σ(-1)^(n-1)/n
=π/-(1/2)ln2
2009-06-20 14:14:55 補充:
π/4-(1/2)ln2