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

Explanation

To find the value of \(\frac{\Sigma_{r=1}^{n} \frac{1}{r}}{\Sigma_{r=1}^{n} \frac{k}{(2n-2k+1)(2n-k+1)}}\), we will break down the problem step by step.

### Step 1: Understand the Summations
We have two summations to evaluate:
1. The numerator: \(\Sigma_{r=1}^{n} \frac{1}{r}\)
2. The denominator: \(\Sigma_{r=1}^{n} \frac{k}{(2n-2k+1)(2n-k+1)}\)

### Step 2: Evaluate the Numerator
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