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2个回答
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绝对收敛,根值审敛法:lim[n->无穷](n^2/4^n)^(1/n)=1/4<1收敛
追问
可不可以把判断过程写一下?谢谢~~
追答
lim[n->无穷](n^2/4^n)^(1/n)=(1/4)lim[n->无穷]n^(2/n)
令lim[n->无穷]n^(2/n)=e^A,
A=ln[lim[n->无穷]n^(2/n)]=lim[n->无穷] (2lnn)/n
罗比达法则=lim[n->无穷] 2/n=0
lim[n->无穷]n^(2/n)=e^0=1
lim[n->无穷](n^2/4^n)^(1/n)=1/4
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