UC知道数列 极限的题目,高手请进!
来源：百度知道 编辑：UC知道 时间：2024/06/01 23:59:53
lim
m→+∞
[(m-1)/1+(m-2)/2+(m-3)/3+....+3/(m-3)+2/(m-2)+1/(m-1)]/m
请问等于多少,给出详细解答,谢谢
先不看极限,+到1/(m-1),比如m是10,就是9/1+8/2+7/3+6/4+5/5+4/6+3/7+2/8+1/9
m→+∞
[(m-1)/1+(m-2)/2+(m-3)/3+....+3/(m-3)+2/(m-2)+1/(m-1)]/m
请问等于多少,给出详细解答,谢谢
先不看极限,+到1/(m-1),比如m是10,就是9/1+8/2+7/3+6/4+5/5+4/6+3/7+2/8+1/9
[(m-1)/1+(m-2)/2+(m-3)/3+....+3/(m-3)+2/(m-2)+1/(m-1)]/m
=[m(1+1/2+1/3+……+1/m)-m]/m
=(1+1/2+1/3+……+1/m)-1
=lnm－1＋欧拉（EULER）常数γ，
近似计算γ=0.5772156.......
lim
m→+∞
[(m-1)/1+(m-2)/2+(m-3)/3+....+3/(m-3)+2/(m-2)+1/(m-1)]/m
=lim 1/m*x从1到(m-1)求和((m-x)/x)
m→+∞
=lim [-1+1/m+x从1到(m-1)求和(1/x)]
m→+∞
=lim [-1+x从1到(m-1)求和(1/x)]
m→+∞
=+∞
=1+1/2+1/3+1/4...1/(m-1)-(1/m+m/1+1/m....1/m)
=1+1/2+1/3+1/4...+1/(m-1)-(m-1)/m
=1/2+1/3+1/4.....+1/m
=∞
因为[(m-1)/1+(m-2)/2+(m-3)/3+....+3/(m-3)+2/(m-2)+1/(m-1)]/m =[(m-1)/1+(m-2)/2+(m-3)/3+....+3/(m-3)+2/(m-2)+1/(m-1)]/m +1-1=1/2+1/3+1/4.....+1/m
所以
请问(m-1)/1+(m-2)/2+.....到哪?+∞ 吗?