Determine if the sequence converges or diverges.
an=(n^4)/(n^3-8n)
We can rewrite the sequence as
an = n(n^3)/(n^3 - 8n)
Dividing both the numerator and denominator by n^3, we get
an = n/(1 - 8/n^2)
As n gets large, the denominator approaches 1, and the entire fraction approaches infinity. Therefore, the sequence diverges to infinity.
To determine if the sequence converges or diverges, we need to find its limit as n approaches infinity.
Let's rewrite the sequence as:
an = (n^4) / (n^3 - 8n)
As n approaches infinity, the terms involving n^3 becomes dominant. We can simplify the expression by dividing both the numerator and denominator by n^3:
an = (n^4 / n^3) / (n^3 / n^3 - 8n / n^3)
= (n / 1) / (1 - 8 / n^2)
Taking the limit of this expression as n approaches infinity:
lim(n→∞) (n / 1) / (1 - 8 / n^2)
Since both the numerator and denominator approach infinity, we can apply L'Hôpital's rule by taking the derivative of the numerator and denominator with respect to n:
lim(n→∞) (1 / 1) / (0 - (-16n) / n^3)
= (1 / 1) / (16n / n^3)
= n^3 / (16n)
Simplifying further:
lim(n→∞) n^2 / 16
As n approaches infinity, the limit becomes infinity divided by a constant, which results in infinity. Therefore, the sequence diverges.