lim (n→oo) [1/(n³+1) +4/(n³+2)+...+n²/ (n³+n)]=? 5
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1^2+2^2+...+n^2 = (1/6)n(n+1)(2n+1)
1/(n³+1) +4/(n³+2)+...+n²/ (n³+n)
≤(1^2+2^2+...+n^2)/(n^3+1)
= (1/6)n(n+1)(2n+1)/(n^3+1)
lim(n->∞) (1/6)n(n+1)(2n+1)/(n^3+1)
=lim(n->∞) (1/6)(1+1/n)(2+1/n)/(1+1/n^3)
=1/3
1/(n³+1) +4/(n³+2)+...+n²/ (n³+n)
≥(1^2+2^2+...+n^2)/(n^3+n)
= (1/6)n(n+1)(2n+1)/(n^3+n)
lim(n->∞) (1/6)n(n+1)(2n+1)/(n^3+n)
=lim(n->∞) (1/6)(1+1/n)(2+1/n)/(1+1/n^2)
=1/3
=>
lim (n->∞) [1/(n³+1) +4/(n³+2)+...+n²/ (n³+n)] =1/3
1/(n³+1) +4/(n³+2)+...+n²/ (n³+n)
≤(1^2+2^2+...+n^2)/(n^3+1)
= (1/6)n(n+1)(2n+1)/(n^3+1)
lim(n->∞) (1/6)n(n+1)(2n+1)/(n^3+1)
=lim(n->∞) (1/6)(1+1/n)(2+1/n)/(1+1/n^3)
=1/3
1/(n³+1) +4/(n³+2)+...+n²/ (n³+n)
≥(1^2+2^2+...+n^2)/(n^3+n)
= (1/6)n(n+1)(2n+1)/(n^3+n)
lim(n->∞) (1/6)n(n+1)(2n+1)/(n^3+n)
=lim(n->∞) (1/6)(1+1/n)(2+1/n)/(1+1/n^2)
=1/3
=>
lim (n->∞) [1/(n³+1) +4/(n³+2)+...+n²/ (n³+n)] =1/3
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