Find the value of `(Sigma_(r=1)^(n) 1/r)/(Sigma_(r=1)^(n) k/((2n-2k+1)(2n-k+1)))`.

Explanation

Let
`A=sum_(k=1)^(n)k/((2n-2k+1)(2n-k+1))`
`=sum_(k=1)^(n)((2n-k+1)-(2n-2k+1))/((2n-2k+1)(2n-k+1))`
`sum_(k=1)^(n)1/(2n-2k+1)-sum_(k=1)^(n)1/(2n-k+1)`
`=(1/1+1/3+1/5+..+1/(2n-1))-(1/(n+1)+1/(n+2)+..+1/(2n))`
and `B=sum_(r=1)^(n)1/r=1/1+1/2+1/3+...+1/n+1/(n+1)+1/(n+2)+..+1/(2n))-(1/1+1/3+1/5+..+1/(2n-1))`
`=1/2+1/4+1/6+...1/(2n)`
`=1/2(1/1+1/2+1/3+..+1/n)`
`=B/2`
So, `B-A=1/2B`
`rArrB/2=A`
`rArrB/A=2`