Is There a Constant Lower Bound for the Integral Test of Convergence?

In summary, the problem is that the first function is smaller than the other one and the second function's integral can diverge. However, if I use a change of variable and solve for ##x>0##, I find that x^5 > x^3+2x^3-1. This shows that the second function is always above the first.
  • #1
Rectifier
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The problem
I am trying to show that the following integral is convergent
$$ \int^{\infty}_{2} \frac{1}{\sqrt{x^3-1}} \ dx $$The attempt
## x^3 - 1 \approx x^3 ## for ##x \rightarrow \infty##.

Since ## x^3 -1 < x^3 ## there is this relation:
##\frac{1}{\sqrt{x^3-1}} > \frac{1}{\sqrt{x^3}}## (the denominator is smaller for function with ## x^3-1 ## and hence bigger quotient).

I know since earlier that the integral ## \int^{\infty}_{2} \frac{1}{\sqrt{x^3}} \ dx ## is convergent. But the problem is that function ##\frac{1}{\sqrt{x^3}}## is smaller than the other one and therefore always "below" that other function. The other functions integral can diverge and ## \int^{\infty}_{2} \frac{1}{\sqrt{x^3}} \ dx ## will still be convergent like nothin' happened.

Please help me to solve this mystery.
 
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  • #2
Why don't you find an integral that is always "above" that other function?
 
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  • #3
fresh_42 said:
Why don't you find an integral that is always "above" that other function?
Will ##\int^{\infty}_{2} \frac{1}{\sqrt{x^3}} + 1## do the job? Smells fishy though.
 
  • #4
Rectifier said:
The problem
I am trying to show that the following integral is convergent
$$ \int^{\infty}_{2} \frac{1}{\sqrt{x^3-1}} \ dx $$The attempt
## x^3 - 1 \approx x^3 ## for ##x \rightarrow \infty##.

Since ## x^3 -1 < x^3 ## there is this relation:
##\frac{1}{\sqrt{x^3-1}} > \frac{1}{\sqrt{x^3}}## (the denominator is smaller for function with ## x^3-1 ## and hence bigger quotient).

I know since earlier that the integral ## \int^{\infty}_{2} \frac{1}{\sqrt{x^3}} \ dx ## is convergent. But the problem is that function ##\frac{1}{\sqrt{x^3}}## is smaller than the other one and therefore always "below" that other function. The other functions integral can diverge and ## \int^{\infty}_{2} \frac{1}{\sqrt{x^3}} \ dx ## will still be convergent like nothin' happened.

Please help me to solve this mystery.

For ##x > 1## we have ##\sqrt{x^3-1} = x^{3/2} \sqrt{1-x^{-3}}##. Try to find a nice constant lower bound on ##\sqrt{1-x^{-3}}## that is good for ##x \geq 2## (or, at least, for large positive ##x##).
 
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  • #5
I don't know. I hope the ##1## isn't meant to be part of the integrand. How about ##x^{-\frac{5}{4}}##?
 
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  • #6
Use a change of variable x = a+1 and (after a little manipulation) compare the integral to the integral of 1/(a3/2)
 
  • #7
FactChecker said:
Use a change of variable x = a+1 and (after a little manipulation)
On which integral?
 
  • #8
Ray Vickson said:
Try to find a nice constant lower bound on ##\sqrt{1-x^{-3}}## that is good for ##x \geq 2## (or, at least, for large positive ##x##).

Why do I need to find a constant lower bound? Isn't the bound x=2 already?
 
  • #9
fresh_42 said:
I don't know. I hope the ##1## isn't meant to be part of the integrand. How about ##x^{-\frac{5}{4}}##?

That one is convergent too but I have to show that it is always above the original function.

In other words:

I must show that
## \frac{1}{\sqrt{x^3-1}}< \frac{1}{x^{\frac{5}{4}}} ## for all x:es

Any suggestions on how I can do that?
 
  • #10
Rectifier said:
That one is convergent too but I have to show that it is always above the original function.

In other words:

I must show that
## \frac{1}{\sqrt{x^3-1}}< \frac{1}{x^{\frac{5}{4}}} ## for all x:es

Any suggestions on how I can do that?
Do some algebra. Remove the quotients, the square roots and so on.
 
  • #11
I solved it another way by showing that the quotient of the first function and the second is not infinite for ## x \rightarrow \infty ##. If you have any alternative solution strategies please share them here.
 
  • #12
I did (##x>0## ) : ##\frac{1}{\sqrt{x^3-1}}<\frac{1}{x^\frac{5}{4}} \Longleftrightarrow x^\frac{5}{4}<\sqrt{x^3-1} \Longleftrightarrow x^\frac{5}{2}<x^\frac{6}{2}-1 \Longleftrightarrow x^6 > x^5 + 2x^3 - 1## which is true for ##x\geq 2## .
 
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  • #13
Rectifier said:
Why do I need to find a constant lower bound? Isn't the bound x=2 already?

There are other ways of doing your problem, but finding a positive constant ##c## such that ##\sqrt{1-x^{-3}} > c## for large ##x>0## is certainly one way to do it. Your previously-expressed worries about ##1/\sqrt{x^3}## not being an upper bound on ##1/\sqrt{x^3-1}## would then go away.
 
Last edited:
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1. What is the integral test of convergence?

The integral test of convergence is a method used to determine the convergence or divergence of an infinite series. It states that if a series can be represented by a continuous, positive, and decreasing function, then the convergence or divergence of the series can be determined by evaluating the corresponding improper integral.

2. How is the integral test of convergence used?

The integral test is used by first finding a continuous, positive, and decreasing function that represents the series. Then, the corresponding improper integral is evaluated. If the integral converges, then the series also converges. If the integral diverges, then the series also diverges.

3. What are the conditions for using the integral test of convergence?

The conditions for using the integral test are that the series must have non-negative terms, the terms must be decreasing, and the series must be continuous. Additionally, the corresponding integral must also be continuous.

4. Are there any limitations to the integral test of convergence?

Yes, the integral test does have limitations. It can only be used for infinite series with non-negative terms and terms that decrease as n increases. It also cannot be used to determine the convergence or divergence of alternating series.

5. Can the integral test of convergence be used to prove absolute convergence?

Yes, the integral test can be used to prove absolute convergence. If the integral converges, then the series also converges absolutely. However, if the integral diverges, the series may still converge conditionally.

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