Question

Let \(a_n=\left(1+\frac{1}{n}\right)^n\) and \(b_n=n\cos(\frac{n!\pi}{2^{10}})\) for n ∈ \(\N\). Then

• A. (a_n) is convergent and (b_n) is bounded
• B. (a_n) is NOT convergent and (b_n) is bounded
• C. (a_n) is convergent and (b_n) is unbounded
• D. (a_n) is NOT convergent and (b_n) is unbounded

Answer 1

The Correct Option is C

To solve the given problem, we need to analyze the behavior of two sequences: \( a_n = \left(1+\frac{1}{n}\right)^n \) and \( b_n = n \cos\left(\frac{n!\pi}{2^{10}}\right) \). We will determine whether \( a_n \) is convergent and whether \( b_n \) is bounded.

1. Analyzing \( a_n = \left(1+\frac{1}{n}\right)^n \):
   • This sequence is known to converge to the number \( e \), which is approximately 2.71828. This result comes from the definition of Euler's number:
   • \(\displaystyle e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n\)
   • As \( n \) approaches infinity, \( a_n \) approaches \( e \). Therefore, \( a_n \) is convergent.
2. Analyzing \( b_n = n \cos\left(\frac{n!\pi}{2^{10}}\right) \):
   • The function \(\cos(x)\) oscillates between -1 and 1 for all \( x \).
   • The term \(\frac{n!\pi}{2^{10}}\) becomes very large for sufficiently large \( n \), leading to essentially random values (since they depend on \( n! \)) when considered with \(\cos\).
   • Therefore, the amplitude of \(\cos\left(\frac{n!\pi}{2^{10}}\right)\) will always be bounded within [-1, 1] but can take these extreme values.
   • Since \( n \) grows indefinitely, while the \(\cos\) term remains bounded, \( b_n \) becomes unbounded as \( n \rightarrow \infty \).

Based on these analyses, the correct answer is that \( a_n \) is convergent to \( e \) and \( b_n \) is unbounded as \( n \) increases. Thus, the correct option is: "(a_n) is convergent and (b_n) is unbounded".