# Convergence/Divergence

by lovelylila
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 P: 17 Determine if the following converges or diverges as x approaches infinity by either evalutation, the direct comparison test, or the limit comparison test: (It's a Calculus II, AP Calculus BC level of problem) the integral of (lnx/(square root of (x^2-1))), from 1 to infinity. * I do not know how to evaluate the integral analytically, so I tried to use either the direct comparison test or limit comparison test. I can't seem to find another function that will "sandwich" that function (and thus prove convergency) or one that will prove it's divergency. I've tried 1/x, 1/(x^2), etc and I'm stuck. Any help on a function to use would be very much appreciated- I'm frustrated beyond belief! Direct Comparison Test: 0< f(x)< g(x) proves that f(x) converges if g(x) also converges f(x)> g(x)---proves that f(x) diverges if g(x) diverges Limit Comparison Test: if the limit as x approaches infinity of f(x)/g(x) is a finite, non-zero number, then f(x) has the same behavior of convergence as g(x)
 HW Helper Sci Advisor P: 2,482 0 < Ln x < x-1 for all x > 1. Does this help?
 P: 39 look at ln(x)/Sqrt(x^2-1) its only improper at infinity so if you look at what the function looks like when x --> infinity we get ln(x) /x . From this we can compare it to anything that diverges and is smaller than that...easiest example 1/x. ln(x)/Sqrt(x^2-1) > 1/x for all x > 1
P: 17

## Convergence/Divergence

thank you, but don't you have to choose a function that is greater/less than for all numbers from 1 to infinity? I don't understand how you can say for x>1, because one itself is the lower limit of the integral and shouldn't it thus be included? or in this type of problem is it to be assumed that the f(x)> g(x) for x>1, not including one?
HW Helper
P: 2,482
That's an easy fix; 0 < Ln x < Sqrt(x-1) for all x > 1. [See edit.]

[Although: "Integral of 1/Sqrt[1+x] does not converge on {1,\[Infinity]}" may be a problem.]

 Quote by Midy1420 ln(x)/Sqrt(x^2-1) > 1/x for all x > 1
"x > 1" should be replaced with "x > 2.5 (approximately)."

I do not think either of these examples is a solution.
 P: 6 thank you all very much for your help , but does anyone know what a solution could be? what you're all suggesting makes sense, but nothing works 100%....
 HW Helper Sci Advisor P: 2,482 On a 2nd thought, 1/x works, but is a little involved. a. For 1 < x < 2.5, 0 < ln(x)/Sqrt(x^2-1) < 1/x. b. For x > 2.5, ln(x)/Sqrt(x^2-1) > 1/x. Fact "a" is inconvenient because it seemingly precludes a direct comparison with 1/x (our hope for a divergence result). So here is the involved part. Below, I will use the expressions "F diverges" and "F = $\infty$" interchangeably. $$\int_1^\infty f(x) dx = \int_1^{2.5} f(x) dx + \int_{2.5}^\infty f(x) dx.$$ In shorthand, I(1/x) = I1(1/x) + I2(1/x) for f(x) = 1/x. 1/x evaluates to a finite (positive) result over [1, 2.5]. Therefore I1(1/x) is finite (and positive). Let I1(1/x) = J > 0. Then, I(1/x) = J + I2(1/x). Since I(1/x) diverges, so does I2(1/x): $\infty$ = J + I2(1/x) implies I2(1/x) = $\infty$ - J = $\infty$. Call this Result 1. From fact "b," I2(ln(x)/Sqrt(x^2-1)) > I2(1/x) = $\infty$ (from Result 1). Therefore I2(ln(x)/Sqrt(x^2-1)) diverges. Call this Result 2. Now... We know we can write I(ln(x)/Sqrt(x^2-1)) = I1(ln(x)/Sqrt(x^2-1)) + I2(ln(x)/Sqrt(x^2-1)). Also, I1(ln(x)/Sqrt(x^2-1)) = K > 0 (positive & finite). Can you take it from here?
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