数列求和:Sn=1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+4+...+n)

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数列求和:Sn=1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+4+...+n)
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数列求和:Sn=1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+4+...+n)
数列求和:Sn=1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+4+...+n)

数列求和:Sn=1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+4+...+n)
分母的通项是an=1+2+...+n=n(n+1)/2
所以Sn=1/a1+1/a2+...+1/an
=2/1*2+2/2*3+...+2/n(n+1)
=2[(1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1))]
=2[1-1/(n+1)]
=2n/(n+1)

Sn=1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+4+...+n)
=1+2[1/(1×2)+1/(2×3)+......+1/(n(n+1))]
=1+2[1-1/(n+1)]
=1+2n/(n+1)

An=1/(1+2+3..+n)
An=1/n(n+1)/2
An=2/n(n+1)