Question

设xn=1/2!+2/3!+····+n/(n+1)!,求n趋向于无穷时xn的极限

Answer 1

1+x/1!+x^2/2!+x^3/3!+x^4/4!+....+x^(n+1)/[(n+1)!]+...=e^x
x/1!+x^2/2!+x^3/3!+x^4/4!+....+x^(n+1)/[(n+1)!]+...=e^x-1

1/1!+x/2!+x^2/3!+x^3/4!+....+x^n/[(n+1)!]+...=(e^x-1)/x
对上式逐项求导得:
1/2!+2x/3!+3x^2/4!+......+nx^(n-1)/[(n+1)!]+...=[(e^x-1)/x]'=[(x-1)e^x+1]/(x^2)
上式中，取x=1，得到所要的结果：
1/2!+2/3!+3/4!+......+n/[(n+1)!]+...=1

Answer 2

n/(n+1)!=1/n!-1/(n+1)!
故xn=∑[1/n!-1/(n+1)!]=1-1/(n+1)!
lim(n--∞)xn=1

Answer 3

xn=1/1!-1/2!+1/3!-1/4!+....+1/n!-1/(n+1)!=1-1/(n+1)!
极限是1

Answer 4

0，这个真不懂

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