Prove that the limit as n approaches infinity of ((n!)^2)/(2n)! is 0.
I'm thinking that you have to use the squeeze theorem, but I'm not quite sure how. help please?
To prove that the limit as n approaches infinity of ((n!)^2)/(2n)! is 0, you can indeed use the squeeze theorem. Let's go through the steps:
Step 1: Express the given expression as a product of two bounds.
We want to find two other expressions that are both greater than or equal to ((n!)^2)/(2n)! and have known limits as n approaches infinity. In this case, we can use the bounds of (1/n)^n and (1/n)^n.
Step 2: Define the lower bound
We define the lower bound as (1/n)^n. This lower bound approaches 0 as n approaches infinity because the exponent becomes larger, making the fraction smaller.
Step 3: Define the upper bound
We define the upper bound as (1/n)^n. This upper bound also approaches 0 as n approaches infinity for the same reason mentioned in Step 2.
Step 4: Use the squeeze theorem
Since the lower bound and upper bound both approach 0 as n approaches infinity, and ((n!)^2)/(2n)! is between the lower and upper bound for all values of n, we can conclude that the limit of ((n!)^2)/(2n)! as n approaches infinity is also 0.
To summarize the proof:
1. Express ((n!)^2)/(2n)! as a product of two bounds, both of which are (1/n)^n.
2. Show that both bounds approach 0 as n approaches infinity.
3. Apply the squeeze theorem to conclude that the limit of ((n!)^2)/(2n)! as n approaches infinity is 0.