Determine whether the following sequences converge or diverge. Justify your answers (show why)
1. a(n) = (e^n)/(3^n)
2. a(n) = (-1)^(2n+1)
3.a(n) = (ln[(e^4)^n])/3n
4. a(n) = (e^n+e^-n)/(e^(2n)-1)
#1 (e/3)^n clearly converges, since e/3 < 1
#2 2n+1 is always odd, so ...
#3 ln[(e^4)^n] = ln[e^(4n)] = 4n ... 4n/3n > 1
#4 e^n + e^-n = (e^2n + 1)/e^n
(e^(2n)+1)/(e^n(e^(2n)-1)) < 1 for n>1
To determine whether the given sequences converge or diverge, we need to analyze their behavior as n approaches infinity. We can do this by finding the limit of the sequence as n approaches infinity. Let's analyze each sequence one by one:
1. a(n) = (e^n)/(3^n)
To determine the behavior of this sequence, we can simplify it as follows:
a(n) = (e/3)^n
As n approaches infinity, the term (e/3)^n goes to zero because 0 < e/3 < 1. Therefore, a(n) converges to 0 as n approaches infinity.
2. a(n) = (-1)^(2n+1)
The sequence a(n) alternates between -1 and 1 with each term. As n approaches infinity, the sequence does not approach a specific value, but rather keeps alternating between -1 and 1. Thus, a(n) diverges as n approaches infinity.
3. a(n) = (ln[(e^4)^n])/3n
We can simplify this sequence as follows:
a(n) = (4n ln e)/(3n)
Since ln e equals 1, the sequence becomes:
a(n) = (4n)/(3n)
As n approaches infinity, the fraction (4n)/(3n) simplifies to 4/3, which is a constant value. Therefore, a(n) converges to 4/3 as n approaches infinity.
4. a(n) = (e^n + e^(-n))/(e^(2n) - 1)
To determine the behavior of this sequence, we can simplify it as follows:
a(n) = (e^n * (1 + e^(-2n)))/(e^(2n) - 1)
As n approaches infinity, e^(-2n) goes to zero, and e^(2n) grows exponentially. Therefore, the term (1 + e^(-2n))/(e^(2n) - 1) approaches 1/2. Multiplying this by e^n, we get a sequence that approaches infinity. Thus, a(n) diverges as n approaches infinity.
In summary:
1. a(n) converges to 0 as n approaches infinity.
2. a(n) diverges as n approaches infinity.
3. a(n) converges to 4/3 as n approaches infinity.
4. a(n) diverges as n approaches infinity.