lim中n→∞n(1-n/1)(1-n/2)...[1-(k-1)/n]=?(求极限limn→∞(n^2)*(k/n -1/(n+1)-

求极限limn→∞(n^2)*(k/n -1/(n+1)-1/(n+2)...-1/(n+k)
求极限 n趋向于无穷大 lim n^2[k/n-1/(n+1)-1/(n+2)-……-1/(n+k)]
求极限lim n→∞(1/(n+1)+1/(n+2)+......+1/(n+n) 求极限(1/(n+1)+1/(n+2)+......+1/(n+n)
lim(n→∞){{n^(n-1)/{[2(n^2)+n+1]^[(n+1)/2]}}}/[1/(n^2)]

lim中n→∞n(1-n/1)(1-n/2)...[1-(k-1)/n]=?(求极限limn→∞(n^2)*(k/n -1/(n+1)-1/(n+2)...-1/(n+k))

求极限limn→∞(n^2)*(k/n -1/(n+1)-1/(n+2)...-1/(n+k)

limn→∞(n^2)*(k/n -1/(n+1)-1/(n+2)...-1/(n+k)

=limn→∞(n^2)*k/n -limn→∞(1/(n+1)+1/(n+2)...+1/(n+k))

其中：当N→-∞或→+∞时，1/(n+1)+1/(n+2)+......+1/(n+k)的极限→0

所以：limn→∞(n^2)*(k/n -1/(n+1)-1/(n+2)...-1/(n+k)

＝limn→∞(nk)

=∞

展开全部

k/n -1 /(n+1) - 1/(n+2)-……-1/(n+k)

= [1/n - 1/(n+1)] + [1/n - 1/(n+2)] + [1/n-1/(n+3)] + ...... + [1/n - 1/(n+k)]

= 1/[n(n+1)] + 2 / [n (n+2)] + 3 /[n(n+3)] + ...... + k / [n(n+k)]

原式 = lim(n->∞) n² { 1/[n(n+1)] + 2 / [n (n+2)] + 3 /[n(n+3)] + ...... + k / [n(n+k)] }

= 1 + 2 + 3 + ...... + k

= k(k+1)/2

函数f(x)=1/(1+x).

用分点将区间[0,1]平均分成n份，分点是

x[k]=k/n,k=1,2,...,n.

利用定积分的定义，和式

∑{f(x[k])*(1/n),k=1...n}

当n->∞时的极限等于定积分

∫{f(x)dx,[0,1]}

而f(x[k])*(1/n)＝1/(n+k)，通项相等，也就是说你的式子等于上面的和式。

于是

lim[1/(n+1) +1/(n+2)+1/(n+3)+……1/(n+n),n->∞]

=∫{f(x)dx,[0,1]}

=∫{1/(1+x)dx,[0,1]}

=ln(1+x)|[0,1]

=ln(1+1)-ln(1+0)

=ln2

原式=lim(n->∞){{n^(n-1)/{[2(n^2)+n+1]^[(n+1)/2]}}}*n^2

=lim(n->∞){{n^(n+1)/{[2(n^2)+n+1]^[(n+1)/2]}}}

=lim(n->∞){n/[2(n^2)+n+1]^0.5}^n+1

lim(n->∞){n/[2(n^2)+n+1]^0.5}<1;

原式=0.