www.whkt.net > limn趋近无穷 1/n^3+(1+2)/n^3+…+(1+2+…+n)/n^3

limn趋近无穷 1/n^3+(1+2)/n^3+…+(1+2+…+n)/n^3

lim (1+2^n+3^n)^1/n = lim (((1/3)^n+(2/3)^n+1)3^n )^1/n= lim 3*((1/3)^n+(2/3)^n+1)^1/n=3

令S(n)=1+(1+2)+(1+2+3)++(1+2++n) 设an = 1+2++n = n(n+1)/2则Sn = a1 + a2 n(n+1)]/2= [(1*1 + 2*2 + + n*n) + (1 + 2 + + n)]/2= [bn + an]/2其中bn = 1*1 + 2*2 +

limn趋近于无穷[(n+1)(n+2)/2]/n^2=limn趋近于无穷(n^2+3n+2)/[2n^2]分子分母同除以n^2所以最后极限就=1/

可以用两边夹法则:因为1^2/(n^3+1)+2^2/(n^3+2)++n^2/(n^3+n) (把分母放缩成n^3+1)=1^2/(n^3+n)+2^2/(n^3+n)++n^2/(n^3+n) =[1/6*n(n+1)(2n+1)]/(n^3+n) (2)注意到n趋于无穷时,(1)(2)两式的极限都是1/3, 所以原式的极限就是1/3.

可以用夹逼定理,n/((n^n+1)^1/n)<(1/n+1+1/(n^2+1)^1/2+…+1/(n^n+1)^1/n)<n/(n+1) 因为lim [n/(n+1)]=lim [n/((n^n+1)^1/n)]=1 所以原极限=1

(1^2+2^2++n^2)/(n^3+n)<=1^2/(n^3+1)+2^2/(n^3+2)+……+n^2/(n^3+n)<=(1^2+2^2++n^2)/(n^3+1),》n(n+1)(2n+1)/6(n^3+n)<=1^2/(n^3+1)+2^2/(n^3+2)+……+n^2/(n^3+n)<=n(n+1)(2n+1)/6(n^3+1),limn→∞ (n+1)(2n+1)/6(n^2+1)=limn→∞ (1+

显然这是n 个数相加,那么小于等于 n/(n^3+1)=1/(n^2+1/n)大于等于 n/(n^3+n) =1/(n^2+1)而在n趋于无穷大的时候,分母都趋于无穷大,那么都趋于0,所以极限值 大于等于0而且小于等于0,故极限值等于0

1+2+3……+2n=(1+2n)*2n/2=n*(1+2n)lim[(1/n^2+1)+(2/n^2+1)+(3/n^2+1)+..+(2n/n^2+1)]lim[(1/n^2+1)+(2/n^2+1)+(3/n^2+1)+..+(2n/n^2+1)]=limit[(2n^2+n)/(n^2+1)]=2

因为:(1+1/n^2)*(1+2/n^2)*(1+3/n^2)*……(1+n/n^2) 评论0 0 0

lim(n->∞){1/(n+1)^2+1/(n+2)^2+……+1/(2n)^2}=∫(0->1) dx/(1+x^2) =arctanx |(0->1)=π/4

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