(n+1)/ （2^n）求和数列求和数列求和0 0

求和【1,+∞】∑(n+1)/2ⁿ
S‹n›=2/2¹+3/2²+4/2³+5/2⁴+6/2⁵+.+n/2ⁿ⁻¹+(n+1)/2ⁿ.(1)
(1/2)S‹n›=2/2²+3/2³+4/2⁴+5/2⁵+6/2⁶+.+n/2ⁿ + (n+1)/2ⁿ⁺¹.(2)
(1)-(2)【错项相减】得：
(1/2)S‹n›=1+1/2²+1/2³+1/2⁴+1/2⁵+1/2⁶+.+1/2ⁿ-(n+1)/2ⁿ⁺¹
=1/2+1/2+1/2²+1/2³+1/2⁴+1/2⁵+1/2ⁿ-(n+1)/2ⁿ⁺¹
=1/2+(1-1/2ⁿ)-(n+1)/2ⁿ⁺¹
故S‹n›=1+2(1-1/2ⁿ)-(n+1)/2ⁿ
故S=n→+∞limS‹n›=n→+∞lim[1+2(1-1/2ⁿ)-(n+1)/2ⁿ]=n→+∞lim{1+2(1-1/2ⁿ)-[1/(2ⁿln2)]}=3.

错位相减。。将Sn*2再减去Sn你就会发现很简单了、、