用放缩法证明1/(1的平方)+1/(2的平方)+1/(3的平方)+......+1/(N的平方)<3/2
应该是这样.谢谢!
...+2\/(2的平方)+3\/(3的平方)+2...+1\/(N的平方)<3\/2
2\/(√k+√k+1) < 2\/(2√k) < 2\/(√k+√k-1)Ⅲ.1\/k^2 的放缩(2)1\/k^2 < 1\/(k^2-1) = 1\/(k+1)(k-1) = (1\/2)[1\/(k-1)-1\/(k+1)]Ⅳ.1\/k^2 的放缩(3)1\/k^2 = 4\/(4k^2) < 4\/(4k^2-1) = 2[1\/(2k-1)-1\/(2k+1)]Ⅴ.变量集中法 |a+b|...
用放缩法证明不等式
<1+(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/(n-1)-1\/n)=2-1\/n<2 所以:1\/(1的平方)+1\/(2的平方)+1\/(3的平方)+...+1\/(N的平方)<2
用放缩法证明1\/(1的平方)+1\/(2的平方)+1\/(3的平方)+...+1\/(N的平方...
当N→+∞时,1\/(1的平方)+1\/(2的平方)+1\/(3的平方)+...+1\/(N的平方)→π^2\/6=1.6449340668482262 >3\/2=1.5
用放缩法证明1\/1^2+1\/2^2+1\/3^2+...+1\/n^2<2(n∈N+)
因为:1\/n<1\/(n-1)(n>1)所以:1\/n^2<1\/n(n-1)=1\/(n-1)-1\/n 因此:1\/1^2+1\/2^2+1\/3^2+...+1\/n^2<1\/1*1+1\/*2+1\/3*2+...+1\/n*(n-1)<1+1-1\/2+1\/2-1\/3...+1\/(n-1)-1\/n=2-1\/n<2
数列的放缩和构造
放缩法:常用于证明数列的不等式,需要注意左右式子的特点,比如有根号,或平方,或有理化。要针对不同的特点来处理,然后再放缩。举个例子:证明(3\/2)*(5\/4)*(7\/6)*…*(2n+1)\/2n>根号n+1,n?正整数,右边有根号,想平方,左边=(3\/2*3\/2)*(5\/4*5\/4)*…*(2n+1\/2n)*(2n+1\/...
用放缩法证明1\/1²+1\/2²+1\/3²+…+1\/n²<2(n∈N+)
①当n=1是,1<2成立 ②当n=2,3,4,……因为1\/n(n-1)>1\/n^2 所以原式<1+1\/2*1+1\/3*2+……+1\/(n-1)n 即 原式<1+1-1\/2+1\/2-1\/3+……+1\/(n-1)-1\/n 即原式<1+1-1\/n 所以原式<2-1\/n<2 故原式<2 由①,②可证 ...
用放缩法证明1\/1²+1\/2²+1\/3²+...+1\/n²<2(n属于N正)
1+1\/2^2+1\/3^2+...+1\/n^2 <1+1\/(1*2)+1\/(2*3)+...+1\/(n-1)n =1+(1-1\/2)+(1\/2-1\/3)+...+(1\/(n-1)-1\/n)=2-1\/n<2 望采纳 O(∩_∩)O谢谢
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!
用放缩法证明 1\/2 - 1\/(n+1) < 1\/(2^2) - 1\/(3^2) + …+ 1\/(n^2...
=(n-1)\/n 所以 1\/(2^2) - 1\/(3^2) + … + 1\/(n^2) < (n-1)\/n (2) 1\/(2^2) - 1\/(3^2) + … + 1\/(n^2)>1\/2*3+1\/3*4+……+1\/n(n+1)=1\/2-1\/3+1\/3-1\/4+……+1\/n-1\/(n+1)=1\/2-1\/(n+1)所以 1\/2 - 1\/(...
用放缩法证明: 1\/2-1\/(n+1)<1\/(2^2)+1\/(3^3)+```+1\/(n^2)<(n-1)\/n...
1\/(2*3)+1\/(3*4)+...+1\/[n(n+1)]=1\/2-1\/3+1\/3-1\/4+...+1\/n-1\/(n+1)=1\/2-1\/(n+1)右半部分 1\/(2^2)+1\/(3^2)+```+1\/(n^2)< 1\/(1*2)+1\/(2*3)+...+1\/[(n-1)n]=1-1\/2+1\/2-1\/3...+1\/(n-1)-1\/n =(n-1)\/n ...