Step 1: Concept Overview:
We will use the Limit Comparison Test to assess the convergence of the given series. This test compares the series with a known convergent or divergent series, such as a p-series.
Step 2: Methodology:
1. Identify the general term, \(a_n\), of the series.
2. Analyze the asymptotic behavior of \(a_n\) for large \(n\). This guides the selection of a comparison series \(b_n\), often a p-series, \(\sum \frac{1}{n^p}\). Recall that a p-series converges if \(p>1\) and diverges if \(p \leq 1\).
3. Implement the Limit Comparison Test: If \(\lim_{n \to \infty} \frac{a_n}{b_n} = L\), where \(L\) is a finite positive number, then \(\sum a_n\) and \(\sum b_n\) share the same convergence/divergence behavior.
Step 3: Detailed Solution:
Given \(a_n = \frac{\sqrt{n+1}+\sqrt{n}}{n^m}\).
For large \(n\), \(\sqrt{n+1}\) approximates \(\sqrt{n}\). Thus, the numerator is approximately \(2\sqrt{n}\).\
\[ a_n \approx \frac{2\sqrt{n}}{n^m} = \frac{2n^{1/2}}{n^m} = \frac{2}{n^{m - 1/2}} \]This suggests comparing with the p-series \(b_n = \frac{1}{n^{m - 1/2}}\).
Applying 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 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 \(L = 2\) is finite and positive, \(\sum a_n\) converges if and only if \(\sum b_n = \sum \frac{1}{n^{m - 1/2}}\) converges.
The p-series \(\sum \frac{1}{n^p}\) converges when \(p>1\). Here, \(p = m - \frac{1}{2}\).\
For convergence:\
\[ m - \frac{1}{2}>1 \]\
\[ m>1 + \frac{1}{2} \]\
\[ m>\frac{3}{2} \]\
Step 4: Conclusion:
The series converges for \(m>\frac{3}{2}\), corresponding to option (D).