Question

The greatest lower bound of the set \[ \left\{ \left( e^n + 2^n \right)^{\frac{1}{n}} : n \in \mathbb{N} \right\}, \] (round off to 2 decimal places) is .............

Hint:

When evaluating limits for sequences involving exponential growth, compare the dominant terms (in this case, \( e^n \) dominates \( 2^n \) for large \( n \)).

Answer 1

Correct Answer: 2.69

Step 1: Analyze the sequence.
The given set is \( S = \left\{ \left( e^n + 2^n \right)^{\frac{1}{n}} : n \in \mathbb{N} \right\} \). We need to find the greatest lower bound (GLB) of this set. We begin by considering the asymptotic behavior of the sequence \( \left( e^n + 2^n \right)^{\frac{1}{n}} \) for large \( n \).
Step 2: Simplify the expression.
As \( n \to \infty \), \( e^n \) grows much faster than \( 2^n \), so we approximate: \[ e^n + 2^n \approx e^n \quad \text{for large} \, n. \] Thus, the sequence behaves like: \[ \left( e^n + 2^n \right)^{\frac{1}{n}} \approx \left( e^n \right)^{\frac{1}{n}} = e. \]
Step 3: Find the limit of the sequence.
To find the GLB, consider the limit of the sequence as \( n \to \infty \): \[ \lim_{n \to \infty} \left( e^n + 2^n \right)^{\frac{1}{n}} = e. \]
Step 4: Conclusion.
Thus, the greatest lower bound of the set is \( e \), which is approximately: \[ \boxed{2.72}. \]