lim[(1/(1+2)+1/(1+2+3)+.+1/(1+2+3+...+n) n趋近于无穷大
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lim[(1/(1+2)+1/(1+2+3)+.+1/(1+2+3+...+n) n趋近于无穷大
![lim[(1/(1+2)+1/(1+2+3)+.+1/(1+2+3+...+n) n趋近于无穷大](/uploads/image/z/18667124-44-4.jpg?t=lim%5B%281%2F%281%2B2%29%2B1%2F%281%2B2%2B3%29%2B.%2B1%2F%281%2B2%2B3%2B...%2Bn%29+n%E8%B6%8B%E8%BF%91%E4%BA%8E%E6%97%A0%E7%A9%B7%E5%A4%A7)
∵1/(1+2+3+...+k)=2/[k(k+1)] (k=1,2,3,.) (应用等差数列求和公式)
=2[1/k-1/(k+1)]
∴1/(1+2)=2(1/2-1/3)
1/(1+2+3)=2(1/3-1/4)
.
1/(1+2+3+.+n)=2[1/n-1/(n+1)]
故 lim(n->∞)[(1/(1+2)+1/(1+2+3)+.+1/(1+2+3+...+n)
=lim(n->∞){2(1/2-1/3)+2(1/3-1/4)+.+2[1/n-1/(n+1)]}
=lim(n->∞){2[1/2-1/(n+1)]}
=2(1/2-0)
=1.
=2[1/k-1/(k+1)]
∴1/(1+2)=2(1/2-1/3)
1/(1+2+3)=2(1/3-1/4)
.
1/(1+2+3+.+n)=2[1/n-1/(n+1)]
故 lim(n->∞)[(1/(1+2)+1/(1+2+3)+.+1/(1+2+3+...+n)
=lim(n->∞){2(1/2-1/3)+2(1/3-1/4)+.+2[1/n-1/(n+1)]}
=lim(n->∞){2[1/2-1/(n+1)]}
=2(1/2-0)
=1.
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