作业帮一个极限求和问题?1/根(2n-1) + 1/[根(2n-1)+根(4n-4)] + 1/[根(4n-4)+根(6n-9)] + ...+ 1/[根（2(i-1)n-(i-1)^2)+根(2in-i^2)] + ...+ 1/[根(2(n-1)*n-(n-1)^2)+根(2n*n-n^2)] 当n趋于无穷大时此式极限为什么为π/4
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/30 18:31:06
![一个极限求和问题?1/根(2n-1) + 1/[根(2n-1)+根(4n-4)] + 1/[根(4n-4)+根(6n-9)] + ...+ 1/[根（2(i-1)n-(i-1)^2)+根(2in-i^2)] + ...+ 1/[根(2(n-1)*n-(n-1)^2)+根(2n*n-n^2)] 当n趋于无穷大时此式极限为什么为π/4](/uploads/image/z/11468432-56-2.jpg?t=%E4%B8%80%E4%B8%AA%E6%9E%81%E9%99%90%E6%B1%82%E5%92%8C%E9%97%AE%E9%A2%98%3F1%2F%E6%A0%B9%282n-1%29+%2B+1%2F%5B%E6%A0%B9%282n-1%29%2B%E6%A0%B9%284n-4%29%5D+%2B+1%2F%5B%E6%A0%B9%284n-4%29%2B%E6%A0%B9%286n-9%29%5D+%2B+...%2B+1%2F%5B%E6%A0%B9%EF%BC%882%28i-1%29n-%28i-1%29%5E2%29%2B%E6%A0%B9%282in-i%5E2%29%5D+%2B+...%2B+1%2F%5B%E6%A0%B9%282%28n-1%29%2An-%28n-1%29%5E2%29%2B%E6%A0%B9%282n%2An-n%5E2%29%5D+%E5%BD%93n%E8%B6%8B%E4%BA%8E%E6%97%A0%E7%A9%B7%E5%A4%A7%E6%97%B6%E6%AD%A4%E5%BC%8F%E6%9E%81%E9%99%90%E4%B8%BA%E4%BB%80%E4%B9%88%E4%B8%BA%CF%80%2F4)
一个极限求和问题?1/根(2n-1) + 1/[根(2n-1)+根(4n-4)] + 1/[根(4n-4)+根(6n-9)] + ...+ 1/[根（2(i-1)n-(i-1)^2)+根(2in-i^2)] + ...+ 1/[根(2(n-1)*n-(n-1)^2)+根(2n*n-n^2)] 当n趋于无穷大时此式极限为什么为π/4
一个极限求和问题?
1/根(2n-1) + 1/[根(2n-1)+根(4n-4)] + 1/[根(4n-4)+根(6n-9)]
+ ...+ 1/[根（2(i-1)n-(i-1)^2)+根(2in-i^2)] + ...+
1/[根(2(n-1)*n-(n-1)^2)+根(2n*n-n^2)]
当n趋于无穷大时此式极限为什么为π/4
一个极限求和问题?1/根(2n-1) + 1/[根(2n-1)+根(4n-4)] + 1/[根(4n-4)+根(6n-9)] + ...+ 1/[根（2(i-1)n-(i-1)^2)+根(2in-i^2)] + ...+ 1/[根(2(n-1)*n-(n-1)^2)+根(2n*n-n^2)] 当n趋于无穷大时此式极限为什么为π/4
引入记号
A=1/{[√(n^2-(n-i+1)^2)]+[√(n^2-(n-i)^2)]}
A1=∑(i从1到n)A
A2=lim(n->∞)A1
B=1/[2√(n^2-(n-i)^2)]
B1=∑(i从1到n)B
B2=lim(n->∞)B1
C=1/[2√(n^2-(n-i+1)^2)]
C1=∑(i从1到n)C,C2=∑(i从1到n+1)C
C3=lim(n->∞)C1,C4=lim(n->∞)C2
因为1/[根（2(i-1)n-(i-1)^2)+根(2in-i^2)]
=1/{[√(n^2-(n-i+1)^2)]+[√(n^2-(n-i)^2)]}
所以 B≤A≤C
B1≤A1≤C1＜C2
B2≤A2≤C3＜C4
又B=1/n[2√(1-(1-i/n)^2)]
所以B2=∫(0到1){1/2√[1-(1-x)^2]}dx(令1-x=sint)=∫(pi/2到0)[-cost/2cost]dt=∫(0到pi/2)1/2dt=pi/4
同理有C3
考虑数列的通项公式为
ai=1/(√((2i-1)n-(i-1)²)+√(2in-i²))
分子分母同时乘以√((2i-1)n-(i-1)²)-√(2in-i²)得
ai=(√((2i-1)n-(i-1)²)-√(2in-i²))/((2i-1)n-(i-1)²-2in+i²)
=(√(...
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考虑数列的通项公式为
ai=1/(√((2i-1)n-(i-1)²)+√(2in-i²))
分子分母同时乘以√((2i-1)n-(i-1)²)-√(2in-i²)得
ai=(√((2i-1)n-(i-1)²)-√(2in-i²))/((2i-1)n-(i-1)²-2in+i²)
=(√((2i-1)n-(i-1)²)-√(2in-i²))/(-2n+2i-1)
分子提出一个i,分母提出一个n得
要转换成定积分的形式吧，这样才能求解
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