第1个回答 2012-09-21
够完整的吧
∑ x^(n+1)/( (n+1)*3^n)
lim n^√[1/(n+1)*3^n]
=1/3
因此收敛域为(-3,3)
记S=∑ x^(n+1)/( (n+1)*3^n)
则S'=∑ (n+1)*x^n/((n+1)*3^n)
=∑ x^n/3^n
=∑(x/3)^n
=1+(x/3)+(x/3)^2+……+(x/3)^n,n趋于无穷
=[1-(x/3)^n]/[1-(x/3)],n趋于无穷
=1/(1-(x/3))
=3/(3-x)
S=S(0)+∫(0,x)S'(y)dy
=∫(0,x)3dy/(3-y)
=(-3)∫(0,x)d(3-y)/(3-y)
=(-3)ln(3-y) | (0,x)
=(-3)ln(3-x)-(-3)ln(3-0)
=(-3)ln(3-x)+3ln3
再讨论端点:
明显x=3,数项级数发散(在S与S'中);x=-3,数项级数也发散(在S'中)
因此,收敛区间为:(-3,3)
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第2个回答 2012-09-21
∑ x^(n+1)/( (n+1)*3^n)
lim n^√[1/(n+1)*3^n]
=1/3
因此收敛域为(-3,3)
记S=∑ x^(n+1)/( (n+1)*3^n)
则S'=∑ (n+1)*x^n/((n+1)*3^n)
=∑ x^n/3^n
=∑(x/3)^n
=1+(x/3)+(x/3)^2+……+(x/3)^n,n趋于无穷
=[1-(x/3)^n]/[1-(x/3)],n趋于无穷
=1/(1-(x/3))
=3/(3-x)
S=S(0)+∫(0,x)S'(y)dy
=∫(0,x)3dy/(3-y)
=(-3)∫(0,x)d(3-y)/(3-y)
=(-3)ln(3-y) | (0,x)
=(-3)ln(3-x)-(-3)ln(3-0)
=(-3)ln(3-x)+3ln3
再讨论端点:
明显x=3,数项级数发散(在S与S'中);x=-3,数项级数也发散(在S'中)
因此,收敛区间为:(-3,3)