✔ 最佳答案
1.lim(n→∞)[n*e^(-n)]=lim(n→∞)(n/e^n)=lim(n→∞)(n)'/(e^n)'.....羅必答法則=lim(n→∞)[1/(n*e^n)]=1/∞=0.....ans
2.lim(n→∞)(-1)^n*(n+3)/(n+2)=lim(n→∞)[+-(n+3)/(n+2)].....even=+, odd=-=lim(n→∞)[+-(n+3)'/(n+2)'].....羅必答法則=lim(n→∞)(+-1/1)=+-1.....ans
2014-02-25 04:38:07 補充：
第一題的第4行是不是有錯><
是不是這樣lim(n→∞)[1/(e^n)](不確定)
Ans:
dn/dn=1
d(e^n)dn=e^n......修正
lim(n→∞)[1/(e^n)]=1/oo=0
2014-02-25 05:14:01 補充：
第2題可否告訴我是收斂還是發散
Ans:
(-1)^n*(n+3)/(n+2)
=(-1)^n*[1+1/(n+2)]
=(-1)^n*1+(-1)^n/(n+2)
=(-1 or 0)+(-1/3+1/4-1/5+1/6-1/7+1/8-+....)
=a+b
a=-1 or 0
b=-1/3+1/4-1/5+1/6-1/7+1/8-+....
=-(1/12+1/30+1/56+.....)
=-Σ1/[(2k+1)(2k+2)].....k=1~n
=-0.5Σ1/[(k+1)(2k+1)]
=-0.5*c
2014-02-25 05:16:07 補充：
c=Σ1/[(k+1)(2k+1)].....Σ->∫as n=oo
=∫dx/(x+1)(2x+1).....x=0~oo
=∫2dx/(2x+1)-∫dx/(x+1)
=∫d(2x+1)/(2x+1)-∫d(x+1)/(x+1)
=ln[(2x+1)/(x+1)]
=ln(2)-ln(3/2)
=ln(2)-ln(3)+ln(2)
=2*ln(2)-ln(3)
b=-0.5c
=0.5*ln(3)-ln(2)
a+b=(-1 or 0)+0.5*ln(3)-ln(2)=收斂.....ans
