Question

Consider the infinite series
\[ (P): \sum_{n=2}^{\infty} \frac{1}{(n \log n)^{1/n}} {and} (Q): \sum_{n=1}^{\infty} \frac{n^n}{(2n)!}. \] Then which one of the following statements is correct?

• A. Series \( (P) \) and \( (Q) \) both converge
• B. Series \( (P) \) converges and series \( (Q) \) diverges
• C. Series \( (P) \) and \( (Q) \) both diverge
• D. Series \( (P) \) diverges and series \( (Q) \) converges

Hint:

To analyze the convergence of series, examine the asymptotic behavior of the general term. If the general term tends to zero too slowly, the series is likely to diverge.

Answer 1

The Correct Option is D

Let's analyze the convergence of both series in detail:
- For series \( (P) \), the general term is \( \frac{1}{(n \log n)^{1/n}} \). To determine whether the series converges, we need to check how the general term behaves as \( n \to \infty \).
- As \( n \) becomes large, \( (n \log n)^{1/n} \) behaves similarly to \( n^{1/n} \), which approaches 1. Thus, the general term behaves like \( \frac{1}{n} \) for large \( n \). Since the series \( \sum \frac{1}{n} \) (the harmonic series) diverges, series \( (P) \) also diverges.
- For series \( (Q) \), the general term is \( \frac{n^n}{(2n)!} \). Using Stirling’s approximation for the factorial, we can approximate the growth of \( (2n)! \). The denominator grows much faster than the numerator, which means the general term decreases very rapidly as \( n \to \infty \). Therefore, the series converges.
Thus, the correct option is (D), where series \( (P) \) diverges and series \( (Q) \) converges.