第四题帮我判断下,12题我就不懂怎么证明了?
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12.
若an有界,又an单调增加,所以an收敛,设an->A
所以存在N>0使n>N时an>A/2
所以n>N时
0<Σ(1-an/an+1) = Σ(1,N)(1-an/an+1) + (a[N+1]-a[N])/a[N] + ...+(a[n+1]-[an])/a[n]
< Σ(1,N)(1-an/an+1) +(a[N+1]-a[N])/(A/2) + ... + (a[n+1]-a[n])/(A/2)
=Σ(1,N)(1-an/an+1) + (-a[N]+a[N+1]-a[N+1]+a[N+2]...-a[n]+a[n+1])/(A/2)
=Σ(1,N)(1-an/an+1) +(a[n+1]-a[N])/(A/2)
n->∞时Σ(1,N)(1-an/an+1) +(a[n+1]-a[N])/(A/2)收敛,故原级数收敛
4.
交错级数,各项绝对值单调减趋于0,故收敛
若an有界,又an单调增加,所以an收敛,设an->A
所以存在N>0使n>N时an>A/2
所以n>N时
0<Σ(1-an/an+1) = Σ(1,N)(1-an/an+1) + (a[N+1]-a[N])/a[N] + ...+(a[n+1]-[an])/a[n]
< Σ(1,N)(1-an/an+1) +(a[N+1]-a[N])/(A/2) + ... + (a[n+1]-a[n])/(A/2)
=Σ(1,N)(1-an/an+1) + (-a[N]+a[N+1]-a[N+1]+a[N+2]...-a[n]+a[n+1])/(A/2)
=Σ(1,N)(1-an/an+1) +(a[n+1]-a[N])/(A/2)
n->∞时Σ(1,N)(1-an/an+1) +(a[n+1]-a[N])/(A/2)收敛,故原级数收敛
4.
交错级数,各项绝对值单调减趋于0,故收敛
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