Word Count : 360

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To solve this numerically and prove the given expression: \[ \lim_{{n \to \infty}} \left[ \frac{1}{\sqrt{2n-1}} + \frac{1}{\sqrt{4n-2^2}} + \frac{1}{\sqrt{6n-3^2}} + \cdots + \frac{1}{n} \right] = \frac{\pi}{2} \] we can break down the summation and analyze its behavior as \( n \to \infty \). ### Step 1: Rewriting the sum The general term in the summation is of the form: \[ \frac{1}{\sqrt{2kn - k^2}} \quad \text{for} \quad k = 1, 2, 3, \dots, n. \] Thus, the sum becomes: \[ S_n = \sum_{k=1}^{n} \frac{1}{\sqrt{2kn - k^2}}. \] For large \( n ________ ________ _____ ______ _____ ___ ___ ____ _______ _________.
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