Question

What is the limit of (n!)^(1/n) as n approaches infinity?

Note: n! means n factorial, which is the product of all positive integers up to n.

Answer 1

Answer:1

Explanation:

Answer 2

Explanation:

To find the limit of (n!)^(1/n) as n approaches infinity, we can use the Stirling's approximation for n!, which is:

n! ≈ (n/e)^n √(2πn)

where e is the mathematical constant e ≈ 2.71828, and π is the mathematical constant pi ≈ 3.14159.

Using this approximation, we can rewrite (n!)^(1/n) as:

(n!)^(1/n) = [(n/e)^n √(2πn)]^(1/n) = (n/e)^(n/n) [√(2πn)]^(1/n)

Taking the limit as n approaches infinity, we have:

lim (n!)^(1/n) = lim (n/e)^(n/n) [√(2πn)]^(1/n)

Using the fact that lim a^(1/n) = 1 as n approaches infinity for any constant a > 0, we can simplify the second term as:

lim [√(2πn)]^(1/n) = 1

For the first term, we can rewrite (n/e)^(n/n) as [1/(e^(1/n))]^n and use the fact that lim a^n = 1 as n approaches infinity for any constant 0 < a < 1. Thus, we have:

lim (n/e)^(n/n) = lim [1/(e^(1/n))]^n = 1

Therefore, combining the two terms, we have:

lim (n!)^(1/n) = lim (n/e)^(n/n) [√(2πn)]^(1/n) = 1 x 1 = 1

Hence, the limit of (n!)^(1/n) as n approaches infinity is 1.