裂项相消问题 1⼀(3눀-1)+1⼀(5눀-1)+1⼀(7눀-1)+…+1⼀[(2n+1)눀-1]+…=?

要求有步骤
2024-12-04 04:14:59
推荐回答(1个)
回答1:


1/[(2n+1)²-1]=1/[(2n+2)(2n)]=1/4[ 1/n-1/(n+1) ]
所以
1/(3²-1)+1/(5²-1)+1/(7²-1)+…+1/[(2n+1)²-1]
=1/4[1-1/(n+1)]
当n向于无穷大时,1/(n+1)=0
所以原式=1/4