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?

• A. \( (x_n) \) and \( (y_n) \) are convergent.
• B. \( (x_n) \) is convergent and \( (y_n) \) is not convergent.
• C. \( (x_n) \) is not convergent and \( (y_n) \) is convergent.
• D. \( (x_n) \) is not convergent and \( (y_n) \) is not convergent.

Hint:

When analyzing convergence of sequences, consider the behavior of the individual terms and compare them with known convergent series or integrals.

Answer 1

The Correct Option is A

Step 1: Look at $x_n$.
Here $x_n = 1+\tfrac12+\cdots+\tfrac1n - \log n$. This is the classic sequence whose limit is the Euler constant $\gamma$.

Step 2: Show $x_n$ settles down.
The gap $x_n - x_{n+1} = \log\!\frac{n+1}{n} - \frac{1}{n+1}$ stays positive and small, and the terms are bounded below by $0$. A bounded monotone sequence converges, so $(x_n)$ has a finite limit.

Step 3: Look at $y_n$.
We have $y_n = \int_1^n \frac{\cos t}{t^2}\,dt$. Compare sizes: $\left|\frac{\cos t}{t^2}\right| \le \frac{1}{t^2}$.

Step 4: The integral converges.
Since $\int_1^{\infty} \frac{1}{t^2}\,dt = 1$ is finite, the integral of $\frac{\cos t}{t^2}$ converges absolutely. So $(y_n)$ also tends to a finite value.

Step 5: Conclusion.
Both sequences converge. So the right choice is the one saying both are convergent, which is option A.
\[ \boxed{\text{Both } (x_n) \text{ and } (y_n) \text{ converge (option A)}} \]