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lim(n→∞){n/(n^2+1)+n/(n^2+2^2)+....+n/(n^2+n^2)}求此式的极限?说明一下那是n除以n的平方加1,
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lim(n→∞)n[1/(n^2+1)+1/(n^2+2)+1/(n^2+3)+……+1/(n^2+n)]
lim(n→∞)[n/(n^2+1)+n/(n^2+2)+n/(n^2+3)+……+n/(n^2+n)]
对于每一项来说,分子均为一次项,分母为二次项,均趋向0
lim(n→∞)n[1/(n^2+1)+1/(n^2+2)+1/(n^2+3)+……+1/(n^2+n)]=0
lim(n→∞)[n/(n^2+1)+n/(n^2+2)+n/(n^2+3)+……+n/(n^2+n)]
对于每一项来说,分子均为一次项,分母为二次项,均趋向0
lim(n→∞)n[1/(n^2+1)+1/(n^2+2)+1/(n^2+3)+……+1/(n^2+n)]=0
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原式=lim(n→∞)[(1/n)/(1+(1/n)^2)+(1/n)/(1+(2/n)^2)+...+(1/n)/(1+(n/n)^2)]
(每项上下同时除以n^2)
=lim(n→∞)1/n*[1/(1+(1/n)^2)+1/(1+(2/n)^2)+...+1/(1+(n/n)^2)]
=∫(0→1)1/(1+x^2)dx
(区间[0,1]的分点为i/n)
=arctanx|(0→1)
=π/4
(每项上下同时除以n^2)
=lim(n→∞)1/n*[1/(1+(1/n)^2)+1/(1+(2/n)^2)+...+1/(1+(n/n)^2)]
=∫(0→1)1/(1+x^2)dx
(区间[0,1]的分点为i/n)
=arctanx|(0→1)
=π/4
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