How would I do this one?
n=1 to infinity n^n/4^1+3*n
Does it absolutely converge, conditionally converge, or diverge?
What would I do here?
n^n /[ 4^(1+3n) ] or what?
Need parentheses to see what you mean.
By the way that clearly does not converge.
This is how it looks: hopefully this clarifies it a bit.
n^n/(4)^1+3*n
but 4^1 is just 4
do you mean
n^n / (4 + 3n) ?
that would be for large n
n^n /3n
(1/3) n^(n-1)
which gets very large indeed
To determine whether the series converges absolutely, conditionally, or diverges, we can use the Ratio Test. The Ratio Test states that if the limit of the absolute value of the ratio of consecutive terms in a series is less than 1, then the series converges absolutely. If the limit is greater than 1 or does not exist, the series diverges. If the limit equals 1, the test is inconclusive.
Let's apply the Ratio Test to the given series:
First, let's find the absolute value of the ratio of consecutive terms:
|a_(n+1) / a_n| = |(n+1)^(n+1) / (4^(1+3n) * n^n)|
Next, we'll simplify this expression:
|a_(n+1) / a_n| = |(n+1)^(n+1) / (4 * n^n * 4^(3n))|
Since we're taking the limit as n approaches infinity, we can focus on the term that grows faster, which in this case is (n+1)^(n+1).
Taking the limit of this term separately:
lim(n → ∞) (n+1)^(n+1) / n^n
We can simplify this limit using exponent rules:
lim(n → ∞) [(n+1) / n]^n * (n+1) / n
Taking the limit as n approaches infinity:
lim(n → ∞) (1 + 1/n)^n * (n+1) / n
This limit is a well-known limit and evaluates to e, Euler's number:
e * (n+1) / n
Now, let's go back to the expression inside the absolute value:
|a_(n+1) / a_n| = |(n+1)^(n+1) / (4 * n^n * 4^(3n))|
Taking the limit as n approaches infinity:
lim(n → ∞) |a_(n+1) / a_n| = e * (n+1) / n * [1 / (4 * 4^(3n))]
The limit can be further simplified:
lim(n → ∞) |a_(n+1) / a_n| = e / 4 * (1 / 4^(3n))
As n approaches infinity, 1 / 4^(3n) approaches 0, and since e / 4 is a constant term, the overall limit of |a_(n+1) / a_n| approaches 0.
Since the limit is less than 1, the series converges absolutely.
Therefore, the given series absolutely converges.