<p>Evaluate \(\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{\sqrt{n(n+k)}}\) [JEE Main 2019]</p>
Step-by-Step Solution
Key Concept: Divide numerator and denominator by n: (1/n) \cdot \Sigma 1/\sqrt{1+k/n} \to \int_0^1 dx/\sqrt{1+x} = 2(\sqrt{2}-1).
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<p>$$\frac{1}{\sqrt{n(n+k)}}=\frac{1}{n\sqrt{1+k/n}}$$</p>
<p>$$\sum_{k=1}^n\frac{1}{\sqrt{n(n+k)}}=\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt{1+k/n}}\to\int_0^1\frac{dx}{\sqrt{1+x}}$$</p>
<p>$$=\left[2\sqrt{1+x}\right]_0^1=2\sqrt{2}-2=2(\sqrt{2}-1)$$</p>
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Correct Answer: A