lim[1/3.5+1/5.7.+1/(2n-1)(2n+1)]
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lim[1/3.5+1/5.7.+1/(2n-1)(2n+1)]lim[1/3.5+1/5.7.+1/(2n-1)(2n+1)]lim[1/3.5+1/5.7.+1/(2n-1)(2n+1)]1/(2
lim[1/3.5+1/5.7.+1/(2n-1)(2n+1)]
lim[1/3.5+1/5.7.+1/(2n-1)(2n+1)]
lim[1/3.5+1/5.7.+1/(2n-1)(2n+1)]
1/(2n-1)(2n+1)=[1/(2n-1)-1/(2n+1)]/2.
所以lim[1/3.5+1/5.7.+1/(2n-1)(2n+1)]
=lim[1/3-1/5+1/5-1/7+1/7-1/9……+1/(2n-1)-1/(2n+1)]/2
=lim[1/3-1/(2n+1)]/2
=lim[1/6-1/(4n+2)]=1/6
裂项相消法:注意到1/(2n-1)(2n+1)=1/(2n-1)-1/(2n+1),
用n=1,2,3,…代入,化简得:原式=lim[1-1/(2n+1)]=1。
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