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I'm studying integral comparison tests and I come to this one.

$$f(x)=\int_{1}^{\infty}\frac{\sqrt{x}}{x^5+\sqrt[3]{x}}dx$$

The solution provided is to do $\displaystyle g(x)=\int_{1}^{\infty}\frac{\sqrt{x}}{x^5}dx$ and use the limit comparison test to find if $f(x)$ is convergent or not.

My question is this. Why can't I use the standard comparison test? Isn't $$0\leq f(x) \leq g(x)$$

Since $g(x)$ is convergent because $\displaystyle \frac{\sqrt{x}}{x^5} = \frac{1}{x^\frac{9}{2}}$ I believe that it's easier to use the standard comparison test and skip the limit comparison test

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You probably have somewhere on your notes that they're equivalent. You can use whatever you want. –  Git Gud Jan 26 '13 at 10:50
    
I think,you mean the integrand is less than or equal to $g(x)$. Isn't it? I mean is $f(x)$ the integrand or $f(x)$ itself is an improper integral? –  B. S. Jan 26 '13 at 10:50
    
@GitGud Hi. No. I only have the LCT. –  Favolas Jan 26 '13 at 10:52
    
@BabakSorouh Sorry. Yes. You're right. It's the integrand –  Favolas Jan 26 '13 at 10:53
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1 Answer 1

up vote 0 down vote accepted

Basically, use any test that works. In this case you are right that the comparison test works and you don't have to use the limit comparison test, though that will work as well. Which one is easier to use, in this case, is a matter of experience and taste.

You need to remember that there are lots of convergence tests (and in general lots of ways to prove any given result) and if you see one used it doesn't mean another can't, or is more difficult, to use. A (very) general rule of thumb for comparison tests is that the limit comparison test is, often, easier to apply. But, as this case shows, not always.

One remark: you should avoid writing things like $f(x)=\int...dx$ since the $x$ on the LHS means something else than the $x$ on the RHS.

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Many thanks for your explanation –  Favolas Jan 26 '13 at 11:19
    
you're welcome! –  Ittay Weiss Jan 26 '13 at 11:27
    
Quick question. I can use the SCT in $\int_{0}^{1}\frac{1}{x^3+\sqrt{x}}dx$ making $g(x)=\int_{0}^{1}\frac{1}{\sqrt{x}}dx$ because $0\leq f(x) \leq g(x)$ Is this safe to assume? –  Favolas Jan 26 '13 at 12:01
    
It is $g(x)=\frac{1}{\sqrt{x}}$. –  André Nicolas Jan 26 '13 at 17:43
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