Question

Consider the two series \[ \text{I. } \sum_{n=1}^{\infty} \frac{1}{n^{1 + (1/n)}} \quad \text{and} \quad \text{II. } \sum_{n=1}^{\infty} \frac{1}{n(\ln n)^{1/n}}. \] Which one of the following holds?

• A. Both I and II converge.
• B. Both I and II diverge.
• C. I converges and II diverges.
• D. I diverges and II converges.

Hint:

Use asymptotic comparisons: if a sequence’s exponent approaches 1 or its logarithmic factor tends to 1, the behavior often mimics the harmonic series, which diverges.

Answer 1

The Correct Option is B

Step 1: Analyze Series I.
\[ a_n = \frac{1}{n^{1 + (1/n)}} = \frac{1}{n \cdot n^{1/n}}. \] Since \(n^{1/n} \to 1\), the general term behaves like \(\frac{1}{n}\). But note that \(n^{1 + 1/n}>n\), so each term is smaller than \(\frac{1}{n}\). To check convergence, compare with \( \sum \frac{1}{n^{1+\varepsilon}} \) where \(\varepsilon>0\). Here the effective exponent \(1 + \frac{1}{n}\) tends to 1, so we can use the integral test: \[ \int_1^\infty \frac{dx}{x^{1+(1/x)}}<\infty. \] Hence, Series I converges slowly but finitely.
Step 2: Analyze Series II.
\[ b_n = \frac{1}{n(\ln n)^{1/n}}. \] For large \(n\), \((\ln n)^{1/n} \to 1\). Thus, \[ b_n \sim \frac{1}{n}. \] Hence, Series II behaves like the harmonic series and diverges.
Step 3: Conclusion.
Series I converges, and Series II diverges. Therefore, (C) is correct.