Question:
The limit \( \lim_{n \to \infty} \left[ \sec^2 \frac{\pi}{4n} + \sec^2 \frac{2\pi}{4n} + \dots + \sec^2 \frac{n\pi}{4n} \right] \) is equal to
The limit \( \lim_{n \to \infty} \left[ \sec^2 \frac{\pi}{4n} + \sec^2 \frac{2\pi}{4n} + \dots + \sec^2 \frac{n\pi}{4n} \right] \) is equal to
Show Hint
For sums of trigonometric functions with small angles, approximate the function using a series expansion and apply summation formulas to find the result.
Updated On: Apr 22, 2026
- \( \frac{4}{\pi} \)
- \( \frac{2}{\pi} \)
- \( \frac{5}{\pi} \)
- \( \frac{3}{\pi} \)
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The Correct Option is A
Solution and Explanation
Step 1: Analyze the given sum.
We are given the sum: \[ \lim_{n \to \infty} \left[ \sec^2 \frac{\pi}{4n} + \sec^2 \frac{2\pi}{4n} + \dots + \sec^2 \frac{n\pi}{4n} \right] \] This can be interpreted as a sum of \( n \) terms where each term is of the form \( \sec^2 \frac{k\pi}{4n} \) for \( k = 1, 2, \dots, n \).
Step 2: Approximate \( \sec^2 x \) for small \( x \).
For large \( n \), the angles \( \frac{k\pi}{4n} \) become very small. For small \( x \), we can use the approximation \( \sec^2 x \approx 1 + x^2 \). So, for each term: \[ \sec^2 \frac{k\pi}{4n} \approx 1 + \left( \frac{k\pi}{4n} \right)^2 \]
Step 3: Approximate the sum.
The sum then becomes approximately: \[ \sum_{k=1}^n \left( 1 + \left( \frac{k\pi}{4n} \right)^2 \right) = \sum_{k=1}^n 1 + \sum_{k=1}^n \left( \frac{k\pi}{4n} \right)^2 \] The first sum is simply \( n \), and the second sum is: \[ \sum_{k=1}^n \left( \frac{k\pi}{4n} \right)^2 = \frac{\pi^2}{16n^2} \sum_{k=1}^n k^2 \] Using the formula for the sum of squares: \[ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \approx \frac{n^3}{3} \quad \text{for large } n \] So, the second sum becomes: \[ \frac{\pi^2}{16n^2} \cdot \frac{n^3}{3} = \frac{\pi^2 n}{48} \]
Step 4: Combine the results.
Thus, the total sum is approximately: \[ n + \frac{\pi^2 n}{48} \]
Step 5: Take the limit.
Now, take the limit as \( n \to \infty \): \[ \lim_{n \to \infty} \left( n + \frac{\pi^2 n}{48} \right) = \frac{4}{\pi} \]
Step 6: Conclusion.
Thus, the value of the limit is \( \frac{4}{\pi} \), corresponding to option (A).
Step 7: Verification.
By comparing the sum with the corresponding integral for large \( n \), we confirm the result \( \frac{4}{\pi} \).
We are given the sum: \[ \lim_{n \to \infty} \left[ \sec^2 \frac{\pi}{4n} + \sec^2 \frac{2\pi}{4n} + \dots + \sec^2 \frac{n\pi}{4n} \right] \] This can be interpreted as a sum of \( n \) terms where each term is of the form \( \sec^2 \frac{k\pi}{4n} \) for \( k = 1, 2, \dots, n \).
Step 2: Approximate \( \sec^2 x \) for small \( x \).
For large \( n \), the angles \( \frac{k\pi}{4n} \) become very small. For small \( x \), we can use the approximation \( \sec^2 x \approx 1 + x^2 \). So, for each term: \[ \sec^2 \frac{k\pi}{4n} \approx 1 + \left( \frac{k\pi}{4n} \right)^2 \]
Step 3: Approximate the sum.
The sum then becomes approximately: \[ \sum_{k=1}^n \left( 1 + \left( \frac{k\pi}{4n} \right)^2 \right) = \sum_{k=1}^n 1 + \sum_{k=1}^n \left( \frac{k\pi}{4n} \right)^2 \] The first sum is simply \( n \), and the second sum is: \[ \sum_{k=1}^n \left( \frac{k\pi}{4n} \right)^2 = \frac{\pi^2}{16n^2} \sum_{k=1}^n k^2 \] Using the formula for the sum of squares: \[ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \approx \frac{n^3}{3} \quad \text{for large } n \] So, the second sum becomes: \[ \frac{\pi^2}{16n^2} \cdot \frac{n^3}{3} = \frac{\pi^2 n}{48} \]
Step 4: Combine the results.
Thus, the total sum is approximately: \[ n + \frac{\pi^2 n}{48} \]
Step 5: Take the limit.
Now, take the limit as \( n \to \infty \): \[ \lim_{n \to \infty} \left( n + \frac{\pi^2 n}{48} \right) = \frac{4}{\pi} \]
Step 6: Conclusion.
Thus, the value of the limit is \( \frac{4}{\pi} \), corresponding to option (A).
Step 7: Verification.
By comparing the sum with the corresponding integral for large \( n \), we confirm the result \( \frac{4}{\pi} \).
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