
n e - ----------- 0 1 1 2 2 2.5 3 2.666666667 4 2.708333333
#include <stdio.h>
double fun ( int i )
{
double re = 2.5, to = 6;
int n;
for ( n = 3; n <= i; n++ )
{
re += 1 / to;
to *= ( n + 1 );
}
return re;
}
int main()
{
int i;
printf ( "n e\n" );
printf ( "- -----------\n" );
printf ( "0 1\n1 2\n2 2.5\n" );
for ( i = 3; i <= 9; i++ )
printf ( "%d %.9lf\n", i, fun ( i ) );
return 0;
}
原文地址:http://blog.csdn.net/acm_baihuzi/article/details/41677307