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 ?
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 ?
\(a_n=p^n n^q(\frac{n}{n+2})^{n^2}\) and \(b_n=\frac{n^n}{n!r^n}(\sqrt{\frac{n+2}{n}})\).
Then, which one of the following statements is TRUE ?
Updated On: Oct 1, 2024
- If 1 < p < e2 and q > 1, then \(\sum\limits_{n=1}^{\infin}a_n\) is convergent
- If e2 < p < e4 and q > 1, then \(\sum\limits_{n=1}^{\infin}a_n\) is convergent
- If 1 < r < e, then \(\sum\limits^{\infin}_{n=1}b_n\) is convergent
- If \(\frac{1}{e}\) < r < e, then \(\sum\limits^{\infin}_{n=1}b_n\) is convergent
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The Correct Option is A
Solution and Explanation
The correct option is (A) : If 1 < p < e2 and q > 1, then \(\sum\limits_{n=1}^{\infin}a_n\) is convergent.
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