一道级数求和问题

an = (3/5)^(n-1)
S(n+1) =a1+a2+...+an+a(n+1) = (5/2) [1- (3/5)^(n+1)]
lim(n->∞) ∑(i:0->n) (3/5)^i
=lim(n->∞) S(n+1)
=lim(n->∞) (5/2) [1- (3/5)^(n+1) ]
=5/2追问

我算的是3/2，首项不是3/5吗

追答

首项
= (3/5)^0
=1

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