How do you determine whether the sequence #a_n=(n^3-2)/(n^2+5)# converges, if so how do you find the limit?

Question

Luke Alvarez · Accepted Answer

No.

If the sequence is convergent then its limit to infinity would be a finite number. #(n^3-2)/(n^2+5)# We can disregard the constants, since this is a limit to infinity.So: #n^3/n^2=n/1# #lim_(n->oo)(n/1)= oo# This doesn't limit to a finite value, therefore it is non-convergent.

Samantha Adkison · Answer

undefined To determine whether the sequence $a_n = \frac{n^3 - 2}{n^2 + 5}$ converges, we examine the behavior of its terms as $n$ approaches infinity. We can use the limit properties to simplify the expression:

$$ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n^3 - 2}{n^2 + 5} $$

We can apply L'Hôpital's Rule to the expression to evaluate the limit:

$$ \lim_{n \to \infty} \frac{n^3 - 2}{n^2 + 5} = \lim_{n \to \infty} \frac{\frac{d}{dn}(n^3 - 2)}{\frac{d}{dn}(n^2 + 5)} $$

$$ = \lim_{n \to \infty} \frac{3n^2}{2n} = \lim_{n \to \infty} \frac{3n^2}{2n} = \lim_{n \to \infty} \frac{3n}{2} = \infty $$

Since the limit of $a_n$ as $n$ approaches infinity diverges to positive infinity, the sequence $a_n$ does not converge.