Question

Let an = \((1+\frac{1}{n})^{\frac{n}{2}}\) be the n^𝑡ℎ term of the sequence 〈a_n 〉, n = 1, 2, 3, …. Then which one of the following is NOT CORRECT?

• A. ⟨an⟩ is bounded
• B. ⟨a_n⟩ is increasing
• C. \(∑^∞ _{n=1} \) in (a_n) is a convergent series
• D. \(lim\\_{n\rightarrow{∞}}(\frac{1}{n}∑^n_{k=1}a_k)=\sqrt{e}\)

Answer 1

The Correct Option is C

To solve this problem, we need to analyze the behavior of the sequence \( \langle a_n \rangle \), given by:

\(a_n = \left(1 + \frac{1}{n}\right)^{\frac{n}{2}}\) 

1. **Bounded Sequence**: We know from calculus that the expression \((1 + \frac{1}{n})^n\) approaches \(e\) as \(n\) approaches infinity. Therefore, \((1 + \frac{1}{n})^{\frac{n}{2}} \approx e^{1/2}\) as \(n\) becomes very large. Because \(a_n\) approaches a finite limit, \( \langle a_n \rangle \) is bounded.
2. **Increasing Sequence**: Consider two consecutive terms of the sequence, \( a_n \) and \( a_{n+1} \). To determine if \( a_n \) is increasing or not, we analyze \( \frac{a_{n+1}}{a_n} \).
3. Simplify:
   • \(a_{n+1} = \left(1 + \frac{1}{n+1}\right)^{\frac{n+1}{2}}\)
   • From this, show that \(\frac{a_{n+1}}{a_n} > 1\) for large \(n\), implying that the sequence is not strictly increasing, and can oscillate initially.
4. **Series Convergence**: For the series \( \sum_{n=1}^{\infty} a_n \) to converge, by the comparison with the harmonic series and knowing that \(a_n \approx e^{1/2}\), we can see that this series does not converge. Therefore, the option about the series being convergent is incorrect.
5. **Limit of Averages**: The expression \(\lim_{n\rightarrow{\infty}} \left(\frac{1}{n} \sum_{k=1}^{n} a_k\right) = \sqrt{e}\) holds true since each \(a_k\) tends towards \(e^{1/2}\) in their limit.

Therefore, the correct answer is that the series \(\sum_{n=1}^{\infty} a_n\) is not convergent, making it the statement that is NOT correct.