Now, let's consider the limit as $n \to \infty$. We can rewrite the sum as an integral:
$\lim_{n \to \infty} \sum_{k=1}^{2n} \frac{1}{n+k} \approx \lim_{n \to \infty} \int_{1}^{2n+1} \frac{1}{n+x} dx$.
Now, we can make the substitution $u = \frac{x}{n}$, so $x
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