In this case, in order to use the comparison theorem to draw a conclusion, we’d have to show that the comparison function ?f(x)? converges. If we find a comparison function that is always greater than the given function, then the given function will be ?g(x)? and the comparison function will be ?f(x)?. Since ?f(x)\geq g(x)? in the comparison theorem: Given an improper integral and asked to use comparison theorem to say whether it converges or diverges, our goal will be to find a comparison function that we know will either always be greater than the given function, or always be less than the given function.
But if ?g(x)? converges, then we can’t draw any conclusion about ?f(x)? because ?f(x)? could diverge or converge above it. If ?g(x)? is less than (below) ?f(x)?, then if ?g(x)? diverges, we know it will force ?f(x)? to also diverge. But if ?f(x)? diverges, then we can’t draw any conclusion about ?g(x)? because ?g(x)? could diverge or converge below it. If ?f(x)? is greater than (above) ?g(x)?, then if ?f(x)? converges, we know it will force ?g(x)? to also converge. The reason is that we’re assuming ?f(x)\geq g(x)?. The comparison theorem will allow you to draw the first two conclusions, but not the others.
If ?\int_a^\infty f(x)\ dx? diverges then so does ?\int_a^\infty g(x)\ dx? If ?\int_a^\infty g(x)\ dx? converges then so does ?\int_a^\infty f(x)\ dx? If ?\int_a^\infty g(x)\ dx? diverges then so does ?\int_a^\infty f(x)\ dx? If ?\int_a^\infty f(x)\ dx? converges then so does ?\int_a^\infty g(x)\ dx?
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