证明:对任何正整数n,都有1+1\/2^2+1\/3^2+……1\/n^2<2
因为n^2大于n(n-1),利用这个放缩,把1\/n^2放大为1\/n(n-1),所以原式小于1+1\/2*1+1\/3*2+...+1\/n(n-1),再利用列项求和的方法,1\/n(n-1)=(1\/n-1)-(1\/n),把所有项都这样写开,发现前后两项可以消掉,最后变为2-1\/n,所以小于2得证。
证明对任意的正整数n,都有1+1\/2^2+1\/3^2+...+1\/n^2≥3n\/2n+1
3n\/2n+1 + 1\/(n+1)^2 >= 3(n+1) \/ (2n+3)即可
证明n为无穷大时,1+1\/2^2+1\/3^2+……+1\/n^2的极限存在
所以原式<1+1\/1-1\/2+1\/2-1\/3+..+1\/(n-1)-1\/n=2-1\/n<2 且原式单增,所以极限存在。
求证1+1\/2^2+1\/3^2十…十1\/n^2<2-1\/n(n∈N,且n≥2)
法二
用放缩法证明1+1\/2^2+1\/3^2+...+1\/n^2<2
1+1\/2*2+1\/3*3+…+1\/n*n< 1+1\/1*2+1\/2*3+…+1\/(n-1)*n=1+1-1\/2+1\/2-1\/3+1\/3…-1\/(n-1)+1\/(n-1)-1\/n=2-1\/n<2 呵呵祝你学习愉快!
证明1+1\/2^2+1\/3^2+...+1\/n^2收敛
证明:因为1+1\/2^2+1\/3^2+...+1\/n^2 < 1+1\/(1*2)+1\/(2*3) +1\/(3*4)+...+ 1\/[(n-1)n]=1+1-1\/2 +1\/2-1\/3+1\/3-1\/4+...+1\/(n-1)-1\/n =2-1\/n <2 有因为 f(n)=1+1\/2^2+1\/3^2+...+1\/n^2 是单调增函数 根据定义,单调增函数 且有上界的...
证明1+1\/2^2+1\/3^2+~~~+1\/n^2<2
1+1\/2²+1\/3²+~~~+1\/n²<1+1\/1·2+1\/2·3+···+1\/[(n-1)n]=1+1\/1-1\/2+1\/2-1\/3+···+1\/(n-1)-1\/n =2-1\/n <2
用数学归纳法证明:1+1\/2^2+1\/3^2+...+1\/n^2<(2n-1)\/n
证明:1+1\/2^2+1\/3^2+...+1\/n^2<(2n-1)\/n (n>=2,n属于N*)1)1+1\/2^2=5\/4 < 3\/2 2) 设:1+1\/2^2+1\/3^2+...+1\/k^2<(2k-1)\/k,1+1\/2^2+1\/3^2+...+1\/k^2+1\/(k+1)^2<(2k-1)\/k+1\/(k+1)^2 =(2k^3+4k^2+2k-k^2-2k-1+k)\/k(k+1...
证明1+1\/2^2+1\/3^2+...+1\/n^2<(2n-1)\/n
证明:1+1\/2^2+1\/3^2+...+1\/n^2<(2n-1)\/n (n>=2,n属于N*)1)1+1\/2^2=5\/4 < 3\/2 2) 设:1+1\/2^2+1\/3^2+...+1\/k^2<(2k-1)\/k,1+1\/2^2+1\/3^2+...+1\/k^2+1\/(k+1)^2<(2k-1)\/k+1\/(k+1)^2 =(2k^3+4k^2+2k-k^2-2k-1+k)\/k(k+1...
证明1+1\/2^2+1\/3^2+...+1\/n^2>3n\/(2n+1)
证明:1+1\/2^2+1\/3^2+...+1\/n^2<(2n-1)\/n (n>=2,n属于N*)1)1+1\/2^2=5\/4 < 3\/2 2)设:1+1\/2^2+1\/3^2+...+1\/k^2<(2k-1)\/k,1+1\/2^2+1\/3^2+...+1\/k^2+1\/(k+1)^2<(2k-1)\/k+1\/(k+1)^2 =(2k^3+4k^2+2k-k^2-2k-1+k)\/k(k+1)...