Word Count : 276

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We are tasked with proving the following limit: \[ \lim_{n\to\infty} \left[ \frac{1}{\sqrt{2n-1^2}} + \frac{1}{\sqrt{4n-2^2}} + \frac{1}{\sqrt{6n-3^2}} + \cdots + \frac{1}{n} \right] = \frac{\pi}{2}. \] ### Step 1: Analyze the General Term The general term in the sum can be written as: \[ \frac{1}{\sqrt{2kn - (k^2)}} \] where \( k = 1, 2, 3, \dots, n ______ __________ _________ ____ __________ __________ _________.
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