Asked by Andy
Determine if the sequence converges or diverges.
an=(n^4)/(n^3-8n)
an=(n^4)/(n^3-8n)
Answers
Answered by
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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.
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.