# How do you use the direct comparison test for improper integrals?

If an improper integral diverges, then we probably do not want to wast time trying to find its value.

Let us assume that we already know:

Let us look examine this uglier improper integral.

By making the numerator smaller and the denominator bigger,

By Comparison Test, we may conclude that

(Intuitively, if the smaller integral diverges, then the larger one has no chance to converge.)

I hope that this was helpful.

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The direct comparison test for improper integrals states that if (f(x)) and (g(x)) are continuous functions for (x \geq a), where (a) is a constant, and (0 \leq f(x) \leq g(x)) for all (x \geq a), then either both integrals (\int_{a}^{\infty} f(x) , dx) and (\int_{a}^{\infty} g(x) , dx) converge or both diverge.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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- How do you determine whether the infinite sequence #a_n=e^(1/n)# converges or diverges?
- How do you determine if the series the converges conditionally, absolutely or diverges given #Sigma ((-1)^(n))/((2n+1)!)# from #[1,oo)#?
- How do you show whether the improper integral #int (x^2)/(9+x^6) dx# converges or diverges from negative infinity to infinity?

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