Question

Let \( a_n = \sqrt{n}, n \geq 1 \), and let \[ s_n = a_1 + a_2 + \cdots + a_n. \text{ Then} \] \[ \lim_{n \to \infty} \left( \frac{a_n}{s_n} \right) \left( -\ln \left( 1 - \frac{a_n}{s_n} \right) \right) = \text{............}. \]

Hint:

For large \( n \), approximate sums by integrals and apply approximations for logarithms when the argument is small.

Answer 1

Correct Answer: 0.9 - 1.1

Step 1: Understanding the series.
We are given that \( a_n = \sqrt{n} \) and \( s_n = \sum_{k=1}^n a_k = \sum_{k=1}^n \sqrt{k} \). We need to evaluate the limit: \[ \lim_{n \to \infty} \left( \frac{a_n}{s_n} \Big/ \ln \left( 1 - \frac{a_n}{s_n} \right) \right) \]
Step 2: Approximation for large \( n \).
For large \( n \), \( a_n = \sqrt{n} \), and the sum \( s_n \) is approximately \( \int_1^n \sqrt{x} dx \), which gives: \[ s_n \approx \frac{2}{3} n^{3/2} \] Thus, the ratio \( \frac{a_n}{s_n} \) is: \[ \frac{a_n}{s_n} \approx \frac{\sqrt{n}}{\frac{2}{3} n^{3/2}} = \frac{3}{2n} \]
Step 3: Applying the logarithm.
Next, we compute \( \ln \left( 1 - \frac{a_n}{s_n} \right) \). Since \( \frac{a_n}{s_n} \) is very small for large \( n \), we use the approximation \( \ln(1 - x) \approx -x \) for small \( x \): \[ \ln \left( 1 - \frac{a_n}{s_n} \right) \approx -\frac{a_n}{s_n} = -\frac{3}{2n} \]
Step 4: Finding the limit.
Now, we substitute this into the original expression: \[ \lim_{n \to \infty} \left( \frac{\frac{3}{2n}}{-\frac{3}{2n}} \right) = \lim_{n \to \infty} 1 = 1 \]
Step 5: Conclusion.
Thus, the value of the limit is \( \boxed{1} \).