Determine if the sequence converges or diverges.
an=e^(-4/(sqrt(n)))
We can use the limit comparison test to compare this sequence to a known sequence with a known convergence/divergence. Let's try comparing it to 1/n:
lim (n → infinity) (e^(-4/sqrt(n)))/(1/n)
= lim (n → infinity) (n/e^(4/sqrt(n)))
= lim (n → infinity) (sqrt(n)/e^(4/sqrt(n)))
We can use L'Hopital's rule to evaluate this limit because it has an infinity/infinity form:
= lim (n → infinity) (1/2e^(4/sqrt(n)))/(1/2n^(-1/2))
= lim (n → infinity) [(n^(-1/2))/(e^(4/sqrt(n)))]
= lim (n → infinity) [(1/(sqrt(n)e^(4/sqrt(n))))]
Since the denominator goes to infinity and the numerator goes to 0, we can conclude that the limit is 0.
Thus, by the limit comparison test, because the limit of (e^(-4/sqrt(n)))/(1/n) is 0 and 1/n converges, we can conclude that the original sequence (an=e^(-4/(sqrt(n)))) also converges.
what is the limit since it converges?
To find the limit of the sequence, we can use the fact that the sequence is monotonically decreasing and bounded below by 0. This implies that the limit exists and is equal to the infimum of the sequence. In other words, we want to find the smallest value that the sequence can get arbitrarily close to.
As n approaches infinity, we can see that the exponent in the sequence, -4/sqrt(n), goes to 0. This means that the terms of the sequence approach e^0=1. Therefore, we have:
lim (n → infinity) e^(-4/(sqrt(n))) = 1.
Hence, the limit of the sequence is 1.
Can you do this one, determine if it converge or diverge,
an= (n^2)e^(-5n)
To determine whether the sequence converges or diverges, we can use the ratio test. We have:
|a_{n+1}/a_n| = |(n+1)^2 e^(-5(n+1))/(n^2 e^(-5n))|
= |(n+1)^2/n^2 e^(-5) e^(-5/n)|
Taking the limit as n approaches infinity:
lim |a_{n+1}/a_n| = lim |(n+1)^2/n^2 e^(-5) e^(-5/n)|
= e^(-5) * lim |(n+1)^2/n^2 e^(-5/n)|
The limit of (n+1)^2/n^2 is 1 as n approaches infinity. To evaluate the limit of e^(-5/n), we can take the natural logarithm and use L'Hopital's rule:
ln lim e^(-5/n) = lim ln(e^(-5/n)) = lim (-5/n)
= 0
Therefore, the limit of e^(-5/n) is 1.
Thus, we have:
lim |a_{n+1}/a_n| = e^(-5)
Since this limit is less than 1, by the ratio test, the series converges. Hence, the sequence an = (n^2)e^(-5n) also converges.
To determine if the sequence converges or diverges, we can examine the behavior of the terms as n approaches infinity (i.e., the limit as n -> infinity).
The given sequence is given by an = e^(-4/√n).
As n approaches infinity, the √n term will also approach infinity, and thus the term 4/√n will approach 0. Therefore, as n approaches infinity, the term e^(-4/√n) will approach e^0, which is equal to 1.
So, the limit of the sequence as n approaches infinity is 1.
Since the limit exists and is finite, we can conclude that the sequence converges.
To determine if the sequence \(a_n=e^{-4/\sqrt{n}}\) converges or diverges, we need to analyze the behavior of the sequence as \(n\) approaches infinity.
First, let's rewrite the sequence as \(a_n = \frac{1}{e^{4/\sqrt{n}}}\).
As \(n\) approaches infinity, the denominator of the fraction approaches infinity because \(\sqrt{n}\) grows without bound.
By using the limit laws, we can rewrite the sequence as:
\[a_n = \frac{1}{e^{\frac{4}{\sqrt{n}}}} = \frac{1}{\left(e^{\frac{1}{\sqrt{n}}}\right)^4}\]
Now, let's evaluate the limit as \(n\) approaches infinity:
\[\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(\frac{1}{e^{\frac{1}{\sqrt{n}}}}\right)^4\]
Since \(\frac{1}{\sqrt{n}}\) approaches \(0\) as \(n\) approaches infinity, we can substitute \(0\) to evaluate the limit:
\[\lim_{n \to \infty} a_n = \left(\frac{1}{e^0}\right)^4 = 1^4 = 1\]
Therefore, the sequence converges to 1 as \(n\) approaches infinity.
In conclusion, the sequence \(a_n = e^{-4/\sqrt{n}}\) converges to 1 as \(n\) approaches infinity.