Question

Consider the series \[\sum_{n=1}^{\infty} n^m \left(\frac{1}{1 + \frac{1}{n^p}}\right)\] where \( m \) and \( p \) are real numbers. Under which of the following conditions does the above series converge?

• A. \( m > 1 \).
• B. \( 0 < m < 1 \) and \( p > 1 \).
• C. \( 0 < m \leq 1 \) and \( 0 \leq p \leq 1 \).
• D. \( m = 1 \) and \( p > 1 \).

Answer 1

The Correct Option is A

To determine the convergence of the series \(\sum_{n=1}^{\infty} n^m \left(\frac{1}{1 + \frac{1}{n^p}}\right)\), we need to analyze its behavior for different values of \( m \) and \( p \).

Step 1: Simplifying the Series Term

The term inside the series can be rewritten and simplified as follows:

\(\frac{1}{1 + \frac{1}{n^p}} = \frac{n^p}{n^p + 1}\).

Thus, the term becomes:

\(\frac{n^m \cdot n^p}{n^p + 1} = \frac{n^{m+p}}{n^p + 1}\).

Step 2: Analyzing the Dominant Behavior

For large values of \( n \), \( n^p + 1 \approx n^p \), so the term can be approximated as:

\(\frac{n^{m+p}}{n^p} = n^m\).

The series approximates to \(\sum_{n=1}^{\infty} n^m\).

Step 3: Applying the p-Series Test

The series \(\sum_{n=1}^{\infty} n^{-a}\) converges if \( a > 1 \). In our case, this is equivalent to checking \(n^{-m}\\) (replacing \( m \) with \( -m \)), implying that convergence occurs if \( -m < -1 \) or \( m > 1 \).

Conclusion

Under the condition \( m > 1 \), the approximation \(\sum_{n=1}^{\infty} n^m\) suggests that the original series converges because for these values, the terms approach zero fast enough.

Therefore, the correct answer is: \( m > 1 \).