怎么判断级数∑(n=1,∞)i^n⼀n是否收敛

2025-03-16 21:20:53
推荐回答(2个)
回答1:

(显然级数不满足绝对收敛,下面判断是否满足条件收敛)

利用欧拉公式:

下面分别讨论实部和虚部的收敛性即可。

当n是奇数时,cos为0;当n是偶数时,sin为0,所以

根据交错级数的莱布尼兹法则,可知实部和虚部都收敛。因此原来的级数收敛。

【纠正一下:倒数第二行,级数的正弦部分应该从n=0开始求和】

回答2:

原级数绝对收敛。

ρ = lim|a/a|

= lim(n+2)! n^(n-1)/[(n+1)^n (n+1)!]

= lim(n+2) n^(n-1)/[(n+1)^n ]

= lim(n+2)/(n+1) lim[n/(n+1)]^(n-1)

= 1* lim{[1-1/(n+1)]^[-(n+1)]}^[-(n-1)/(n+1)]

= e^lim -(n-1)/(n+1) = e^lim -(1-1/n)/(1+1/n) = 1/e < 1.

扩展资料

1.发散与收敛对于数列和函数来说,它就只是一个极限的概念,一般来说如果它们的通项的值在变量趋于无穷大时趋于某一个确定的值时这个数列或是函数就是收敛的,所以在判断是否是收敛的就只要求它们的极限就可以了.对于证明一个数列是收敛或是发散的只要运用书上的定理就可以了。

2.对于级数来说,它也是一个极限的概念,但不同的是这个极限是对级数的部分和来说的,在判断一个级数是否收敛只要根据书上的判别法就行了。

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