Determine whether the infinite sequence converges or diverges. If it converges, find the limit.
{(1)(3)(5)...(2n-1) / (2n)^n}
For n > 2, you have
1*3*5/6*6*6 < 3/6^2
1*3*5*7/8*8*8*8 < 3/8^2
and so on. The nth term is less than 3/(2n)^2 so it converges to 0
To determine whether the infinite sequence converges or diverges, we need to analyze the behavior of each term in the sequence as n approaches infinity.
Let's start by simplifying the expression:
{(1)(3)(5)...(2n-1) / (2n)^n}
We can rewrite the numerator as (2n-1)!! to represent the product of all odd numbers from 1 to (2n-1). Similarly, we can rewrite the denominator as (2n)^n.
Now, we can rewrite the sequence as:
((2n-1)!!) / ((2n)^n)
Since the terms involve factorials, it's helpful to apply the ratio test, which states that if the limit of the absolute value of the ratio of consecutive terms is less than 1, then the series converges. Otherwise, if the limit is greater than 1 or approaches infinity, the series diverges.
Let's apply the ratio test to our sequence:
lim(n->∞) |((2(n+1)-1)!! / ((2(n+1))^n) * ((2n)^n)/((2n-1)!!)|
We can simplify this limit:
lim(n->∞) |((2n+1)!! / (2n)!) * (2n)^n / ((2n-1)!!)|
Next, let's use the properties of factorials:
- (2n+1)!! = (2n+1) * (2n-1) * (2n-3) * ... * 3 * 1
- (2n)!! = (2n) * (2n-2) * (2n-4) * ... * 4 * 2
- (2n-1)!! = (2n-1) * (2n-3) * (2n-5) * ... * 3 * 1
- (2n-2)!! = (2n-2) * (2n-4) * (2n-6) * ... * 4 * 2
Simplifying the ratio further, we get:
lim(n->∞) |[1/(2n+1)] * [(2n)^n] / [1/(2n-1)]|
After canceling out similar terms, we have:
lim(n->∞) [1/(2n+1)] * [(2n)^n] * [(2n-1)]
Now, let's simplify the limit using properties of exponents:
lim(n->∞) [1/(2n+1)] * [(2n)^n] * [(2n-1)]
= [1/(2n+1)] * [(2n)^(2n/2)] * [(2n-1)]
= [1/(2n+1)] * [(2n)^2n] * [(2n-1)]
= [1/(2n+1)] * [2^(2n)] * [n^(2n)] * [(2n-1)]
Now, we can take the limit as n approaches infinity:
lim(n->∞) [1/(2n+1)] * [2^(2n)] * [n^(2n)] * [(2n-1)]
= 0 * ∞ * ∞ * ∞
Since we have an indeterminate form, further analysis is required. Let's apply L'Hopital's rule by taking the natural logarithm of the expression:
lim(n->∞) ln([1/(2n+1)] * [2^(2n)] * [n^(2n)] * [(2n-1)])
= lim(n->∞) ln[1/(2n+1)] + ln[2^(2n)] + ln[n^(2n)] + ln[(2n-1)]
Using logarithm properties, this becomes:
= lim(n->∞) ln[1/(2n+1)] + 2n*ln[2] + 2n*ln[n] + ln[(2n-1)]
Now, we can apply L'Hopital's rule to each term of the sum:
= lim(n->∞) [-(2)/(2n+1)] + 2ln[2] + 2ln[n] + [-(2)/(2n-1)]
Taking the limit, we find that all these terms approach zero except for 2ln[2]:
= 0 + 2ln[2] + 0 + 0
= 2ln[2]
Thus, the limit of the sequence is 2ln[2] as n approaches infinity.
Therefore, the infinite sequence converges, and its limit is 2ln[2].