Question

For p, q, r ∈ ℝ, r ≠ 0 and n ∈ \(\N\), let
\(a_n=p^n n^q(\frac{n}{n+2})^{n^2}\) and \(b_n=\frac{n^n}{n!r^n}(\sqrt{\frac{n+2}{n}})\).
Then, which one of the following statements is TRUE ?

• A. If 1 < p < e² and q > 1, then \(\sum\limits_{n=1}^{\infin}a_n\) is convergent
• B. If e² < p < e⁴ and q > 1, then \(\sum\limits_{n=1}^{\infin}a_n\) is convergent
• C. If 1 < r < e, then \(\sum\limits^{\infin}_{n=1}b_n\) is convergent
• D. If \(\frac{1}{e}\) < r < e, then \(\sum\limits^{\infin}_{n=1}b_n\) is convergent

Answer 1

The Correct Option is A

- For \( a_n = p^n n^q \left( \frac{n}{n+2} \right)^{n^2} \): The term \( \left( \frac{n}{n+2} \right)^{n^2} \) behaves like \( e^{-2n} \) for large \( n \). Therefore, for large \( n \), we have: \[ a_n \sim p^n n^q e^{-2n}. \] This expression decays exponentially due to the \( e^{-2n} \) term, but it also grows polynomially with \( n^q \). The series \( \sum_{n=1}^{\infty} a_n \) converges if the exponential decay dominates the polynomial growth. The critical condition for convergence is \( 1 < p < e^2 \), which ensures that the exponential term decays sufficiently fast. Additionally, \( q > 1 \) ensures that the polynomial term does not cause divergence. - For \( b_n = \frac{n^n}{n!} r^n \left( \sqrt{\frac{n+2}{n}} \right) \): The term \( \frac{n^n}{n!} \) grows very quickly for large \( n \), and the factor \( r^n \) doesn't decay fast enough to counter this rapid growth. Therefore, \( \sum_{n=1}^{\infty} b_n \) diverges for typical values of \( r \). Thus, the correct answer is (A): If \( 1 < p < e^2 \) and \( q > 1 \), then \( \sum_{n=1}^{\infty} a_n \) is convergent