Question:
For each positive integer \( n \), let
\[
x_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \log n
\]
and
\[
y_n = \int_1^n \frac{\cos t}{t^2} \, dt.
\]
Which ONE of the following statements about the sequences \( (x_n) \) and \( (y_n) \) is TRUE?
For each positive integer \( n \), let
\[
x_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \log n
\]
and \[ y_n = \int_1^n \frac{\cos t}{t^2} \, dt. \]
Which ONE of the following statements about the sequences \( (x_n) \) and \( (y_n) \) is TRUE?
and \[ y_n = \int_1^n \frac{\cos t}{t^2} \, dt. \]
Which ONE of the following statements about the sequences \( (x_n) \) and \( (y_n) \) is TRUE?
Show Hint
When analyzing convergence of sequences, consider the behavior of the individual terms and compare them with known convergent series or integrals.
Updated On: Jun 1, 2026
- \( (x_n) \) and \( (y_n) \) are convergent.
- \( (x_n) \) is convergent and \( (y_n) \) is not convergent.
- \( (x_n) \) is not convergent and \( (y_n) \) is convergent.
- \( (x_n) \) is not convergent and \( (y_n) \) is not convergent.
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The Correct Option is A
Solution and Explanation
Step 1: Analyze \( x_n \).
The sequence \( x_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \log n \) is the sum of the harmonic series minus the logarithmic term.
The harmonic series diverges, but the subtraction of \( \log n \) ensures that \( x_n \) converges to a finite limit. Thus, \( (x_n) \) is convergent.
Step 2: Analyze \( y_n \).
The integral \( y_n = \int_1^n \frac{\cos t}{t^2} \, dt \) represents a function that decays as \( t \) increases because \( \frac{\cos t}{t^2} \) behaves similarly to \( \frac{1}{t^2} \), which converges to zero.
Since the integrand decays rapidly, the integral converges to a finite value. Therefore, \( (y_n) \) is convergent.
Step 3: Conclusion.
From the analysis, both sequences \( (x_n) \) and \( (y_n) \) are convergent. Therefore, the correct answer is (C).
The sequence \( x_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \log n \) is the sum of the harmonic series minus the logarithmic term.
The harmonic series diverges, but the subtraction of \( \log n \) ensures that \( x_n \) converges to a finite limit. Thus, \( (x_n) \) is convergent.
Step 2: Analyze \( y_n \).
The integral \( y_n = \int_1^n \frac{\cos t}{t^2} \, dt \) represents a function that decays as \( t \) increases because \( \frac{\cos t}{t^2} \) behaves similarly to \( \frac{1}{t^2} \), which converges to zero.
Since the integrand decays rapidly, the integral converges to a finite value. Therefore, \( (y_n) \) is convergent.
Step 3: Conclusion.
From the analysis, both sequences \( (x_n) \) and \( (y_n) \) are convergent. Therefore, the correct answer is (C).
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