数列n+1的阶乘分之n求和

阮洁仪3866求高手给个1到N各数阶乘分之一和的C语言算法 -
荀饺垄13049536882 ______ #include<stdio.h> #include<math.h> int main() { int i,j,n,s1; double s; scanf(＂%d＂,&n); for(i=1,s=0;i<=n;i++) { for(j=1,s1=1;j<=i;j++) s1=s1*j; s=s+1.0/s1; } printf(＂%f＂,s); return 0; }

阮洁仪3866求数列{n(n+1)/1}的前n项和Sn=?就是{n(n+1)分之1} -
荀饺垄13049536882 ______[答案] 1/n(n+1)=1/n-1/(n+1) Sn=1-1/2+1/2-1/3+...+1/(n-1)-1/n+1/n-1/(n+1) =1-1/(n+1) =n/(n+1)

阮洁仪3866求数列的和 an=n乘n的阶乘求sn -
荀饺垄13049536882 ______[答案] 因为 an=n*n!=[(n+1)-1]*n!=(n+1)!-n! , 所以 Sn=a1+a2+.+an =(2!-1!)+(3!-2!)+.+[(n+1)!-n!] =(n+1)!-1 .

阮洁仪3866n趋近于无穷时,n的阶乘的1/n除以n的极限怎么求? -
荀饺垄13049536882 ______[答案] 答案是1/e.令xn=((n+1)/e)∧n,则xn/xn-1=(1/e)*(1+1/n)∧n*n

阮洁仪3866数学,已知数列n(n+1)分之1 求前n项和Sn -
荀饺垄13049536882 ______ an=1/n(n+1)=1/n-1/(n+1) Sn=1-1/2+1/2-1/3+...+1/n-1/(n+1) =1-1/(n+1) =n/(n+1)

阮洁仪3866已知数列n(n+1)分之1 求前n项和Sn -
荀饺垄13049536882 ______[答案] an=1/n(n+1)=1/n-1/(n+1) Sn=1-1/2+1/2-1/3+...+1/n-1/(n+1) =1-1/(n+1) =n/(n+1)

阮洁仪3866已知数列通项an=n(n+1)分之1求数列前n项和sn -
荀饺垄13049536882 ______ an=1/[n(n+1)]=(n+1-n)/[n(n+1)]=1/n-1/(n+1) Sn=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+[1/(n-1)-1/n]+[1/n-1/(n+1)] =1-1/(n+1) =n/(n+1) 您的采纳是我们的动力.

阮洁仪3866正整数从1到n的连乘积叫做n的阶乘,.已知数列an的通项公式an=log(n+1)(2008!),则1/a1/+1/a2````+1/a2007= -
荀饺垄13049536882 ______[答案] an=log(n+1)(2008!) 1/an=log(2008!)(n+1) 1/a1+1/a2+...+1/a2007 =log(2008!)(1+1)+log(2008!)(2+1)+...+log(2008!)(2007+1) =log(2008!)(2)+log(2008!)(3)+...+log(2008!)(2008) =log(2008!)(2*3*...*2008) =log(2008!)(1*2*3*...*2008) =log(2008!)(2008!) =1

阮洁仪3866高手如何证明这个数列是发散的【用反证法】证明:数列{Un}={( - 1)的n次方乘以n+1分之n} -
荀饺垄13049536882 ______[答案] 假设数列{Un}是收敛数列}(-1)的n次方乘以n+1分之n的极限为a由于(-1)的n次方*n/n+1=(-1)的n次方-(-1)的n次方/n+1因为(-1)的n次方/n+1的极限为0(n趋于无穷时)则数列{(-1)的n次方/n+1}收敛则 (-1)的n次方的极限=(...

阮洁仪38661的阶乘+2倍2的阶乘+.+n倍n的阶乘=? -
荀饺垄13049536882 ______[答案] n*n!=[(n+1)-1]*n!=(n+1)!-n! 1*1!=2!-1! 2*2!=3!-2! …… 1的阶乘+2倍2的阶乘+.+n倍n的阶乘=(2!-1!) +(3!-2!) +(4!-3!) +……+[(n+1)!-n!]=(n+1)!-1!=(n+1)!-1