数列(1+1/n)的(n+p)次方严格递减,p的取值范围是多少
时间: 2024-01-29 08:05:05 浏览: 29
首先,我们可以将数列(1+1/n)的(n+p)次方展开,得到下列结果:
(1+1/n)^(n+p) = 1 + (n+p)/(1!) * (1/n) + (n+p)(n+p-1)/(2!) * (1/n)^2 + ... + (n+p)(n+p-1)...(n+1)/(p!) * (1/n)^p
我们注意到,对于上式中的任意一项,它的系数都是大于0的。而且,当n趋近于正无穷时,这些项的系数都趋近于1。因此,我们只需考虑这些项的次数对数列递减的影响。
具体地,我们可以将相邻两项之间的差值表示为:
(1+1/n)^(n+p) - (1+1/(n+1))^(n+p) = [(n+p)/(1!) * (1/n) + (n+p)(n+p-1)/(2!) * (1/n)^2 + ... + (n+p)(n+p-1)...(n+2)/(p!) * (1/n)^(p-1) + (n+p)(n+p-1)...(n+1)/(p!) * (1/n)^p] - [(n+p)/(1!) * (1/(n+1)) + (n+p)(n+p-1)/(2!) * (1/(n+1))^2 + ... + (n+p)(n+p-1)...(n+2)/(p!) * (1/(n+1))^(p-1) + (n+p)(n+p-1)...(n+1)/(p!) * (1/(n+1))^p]
将上式中的所有项通分,并化简,可以得到:
(1+1/n)^(n+p) - (1+1/(n+1))^(n+p) = [(n+p)/(1!) * (1/n) - (n+p)/(1!) * (1/(n+1))] + [(n+p)(n+p-1)/(2!) * (1/n)^2 - (n+p)(n+p-1)/(2!) * (1/(n+1))^2] + ... + [(n+p)(n+p-1)...(n+2)/(p!) * (1/n)^(p-1) - (n+p)(n+p-1)...(n+2)/(p!) * (1/(n+1))^(p-1)] + [(n+p)(n+p-1)...(n+1)/(p!) * (1/n)^p - (n+p)(n+p-1)...(n+1)/(p!) * (1/(n+1))^p]
我们发现,上式中每一对方括号内的差值都是大于0的,因此整个差值也是大于0的。这意味着数列(1+1/n)的(n+p)次方是严格递减的。
综上所述,p的取值范围可以是任意正实数。