Question

求极限lim n→∞ (2^n)/n!

稍微详细一点

Answer 1

对于级数∑(2^n)/n!
令Un=(2^n)/n!
u(n+1)=(2^(n+1))/(n+1)!
lim(n→∞)u(n+1)/un
=lim(n→∞)[(2^(n+1))/(n+1)!]/[(2^n)/n!]
=lim(n→∞)2/(n+1)
=0
即级数收敛，所以
它的一般项极限=0

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∑(2^n)/n!的和怎么求

Answer 2

lim [n√(2^n+3^n+5^n)]e^{lim [(1/n)*ln(2^n+3^n+5^n)]}
对lim [(1/n)*ln(2^n+3^n+5^n)]用L'HOPITAL法则

lim [(1/n)*ln(2^n+3^n+5^n)]
=lim [(ln2*2^n+ln3*3^n+ln5*5^n)/(2^n+3^n+5^n)]
=lim {[ln2*(2/5)^n+ln3*(3/5)^n+ln5]/[(2/5)^n+(3/5)^n+1]}
当n→∞时(2/5)^n和(3/5)^n均趋向于0
故lim [1/n *ln(2^n+3^n+5^n)] = ln5

lim (n√2^n+3^n+5^n)
=e^{lim [(1/n)*ln(2^n+3^n+5^n)]}
=e^ln5
=5

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额，你是复制粘贴的么，明显不是同一个题目