Respuesta :
The given sequence is convergent and tends to zero.
Sequence:
In real analysis, a sequence is a function from the set of natural numbers N to the set of real numbers R.
We can say that a sequence is a function f: N-->R.
Convergent sequences:
- A sequence is said to be convergent if it approaches a finite number.
- A sequence [tex]a_{n}[/tex] having 'n' terms is said to be convergent if its limit exists.
Divergent sequences:
- A sequence is said to be divergent if it doesn't approach a finite number.
- The limit of the sequence does not exist.
Factorial:
- A factorial of a number n is the product of all the numbers ≤ n
It is calculated as n! = 1.2.3....n.
The given sequence is (3n − 1)!/ (3n + 1)!
Let [tex]a_{n}[/tex] = (3n − 1)!/ (3n + 1)! .
(3n-1)! = 1.2.3.....(3n-1)
(3n + 1)! = 1.2.3.....(3n-1)(3n)(3n+1)
So the given sequence becomes
[tex]a_{n}[/tex] = [tex]\frac{ 1.2.3.....(3n-1)}{1.2.3.....(3n-1)(3n)(3n+1)}[/tex]
⇒ [tex]a_{n}[/tex] = [tex]\frac{1}{(3n)(3n+1)}[/tex] = [tex]\frac{1}{9n^{2} +3n}[/tex].
As n tends to infinity, the value of 9n²+3n increases. So the value [tex]\frac{1}{9n^{2} +3n}[/tex] tends to zero.
[tex]\lim_{n \to \infty} a_n[/tex] = 0.
Hence, the given sequence converges and tends to zero.
The correct question is:
Determine whether the sequence converges or diverges. If it converges, find the limit. (If an answer does not exist, enter DNE.) (3n − 1)! /(3n + 1)!
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