Question

The values of 'm' for which the infinite series,
\(\sum \frac{\sqrt{n+1}+\sqrt{n}}{n^m}\) converges, are:

• A. \(m>\frac{1}{3}\)
• B. \(m>\frac{1}{2}\)
• C. \(m>1\)
• D. \(m>\frac{3}{2}\)

Hint:

When analyzing the convergence of a series with a complex term, first find its "dominant" behavior for large \(n\). In this case, \(\sqrt{n+1}+\sqrt{n}\) behaves like \(2\sqrt{n}\). This simplifies the term to \(2/n^{m-1/2}\), immediately identifying it as a p-series and making the condition for convergence easy to find.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To determine the convergence of the given series, we can use the Limit Comparison Test. This test compares the given series with a known convergent or divergent series (like a p-series) to determine its behavior.

Step 2: Key Formula or Approach:
1. Identify the general term of the series, \(a_n\).
2. Determine the asymptotic behavior of \(a_n\) for large \(n\). This helps in choosing a suitable series \(b_n\) for comparison. A p-series, \(\sum \frac{1}{n^p}\), is often used, which converges for \(p>1\) and diverges for \(p \leq 1\).
3. Apply the Limit Comparison Test: If \(\lim_{n \to \infty} \frac{a_n}{b_n} = L\), where \(L\) is a finite positive constant, then \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge.

Step 3: Detailed Explanation:
Let the general term of the series be \(a_n = \frac{\sqrt{n+1}+\sqrt{n}}{n^m}\).
For large \(n\), \(\sqrt{n+1} \approx \sqrt{n}\). So, the numerator behaves like \(\sqrt{n} + \sqrt{n} = 2\sqrt{n}\). \[ a_n \approx \frac{2\sqrt{n}}{n^m} = \frac{2n^{1/2}}{n^m} = \frac{2}{n^{m - 1/2}} \] This suggests we should compare our series with the p-series \(b_n = \frac{1}{n^{m - 1/2}}\).
Let's apply the Limit Comparison Test: \[ L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{\sqrt{n+1}+\sqrt{n}}{n^m}}{\frac{1}{n^{m - 1/2}}} \] \[ L = \lim_{n \to \infty} \frac{(\sqrt{n+1}+\sqrt{n}) . n^{m - 1/2}}{n^m} \] \[ L = \lim_{n \to \infty} \frac{\sqrt{n+1}+\sqrt{n}}{n^{1/2}} \] Divide the numerator and denominator by \(\sqrt{n}\): \[ L = \lim_{n \to \infty} \left( \frac{\sqrt{n+1}}{\sqrt{n}} + \frac{\sqrt{n}}{\sqrt{n}} \right) = \lim_{n \to \infty} \left( \sqrt{\frac{n+1}{n}} + 1 \right) \] \[ L = \lim_{n \to \infty} \left( \sqrt{1 + \frac{1}{n}} + 1 \right) = \sqrt{1+0} + 1 = 2 \] Since the limit \(L = 2\) is a finite and positive number, the series \(\sum a_n\) converges if and only if the series \(\sum b_n = \sum \frac{1}{n^{m - 1/2}}\) converges.
The p-series \(\sum \frac{1}{n^p}\) converges when its exponent \(p\) is greater than 1. In our case, \(p = m - \frac{1}{2}\). So, for convergence, we must have: \[ m - \frac{1}{2}>1 \] \[ m>1 + \frac{1}{2} \] \[ m>\frac{3}{2} \]
Step 4: Final Answer:
The series converges for \(m>\frac{3}{2}\), which corresponds to option (D).