## anonymous 5 years ago Comparison test to determine if improper integral converges or diverges, (3+cos(x))/(x^3) from 1 to infinity

1. anonymous

3+cos(x) varies between 4 and 2 as x approaches infinity. Since 4 >= 3+cos(x) always, the series of (3+cos(x))/x^3 will converge if 4/x^3 converges.

2. anonymous

f(x) >= g(x) >= 0 f(x) = 1/x^2 g(x) = (3+cos(x))/(x^3) because of f(x) is larger for all values of g(x) and f(x) converges to $\int\limits_{1}^{∞} 1/x^2 dx = \lim t \rightarrow∞ \int\limits_{1}^{t} 1/x^2 = {-1/x}$ from t to1 = 1

3. anonymous

1/x^2 is not larger than (3+cos(x))(x^3) For quite a few values greater than 1.

4. anonymous

Infinitely many in fact.

5. anonymous

I dont understand how he got the 1/x^2 :P , also whats the difference if you select 2 instead of 4? say 2<=3+cos(x)?

6. anonymous

You want to select something that is greater than the function you want to solve. If you can find something which converges but which is strictly larger than the function you're looking to solve, then you know that your function must also converge.

7. anonymous

Oh ok, so when I get 4/x^3 do I have to evaluate the definite integral of that and find out if it converges or diverges?

8. anonymous

Oops i meant (3+cos(x))/ x^2 which is greater then all values of (3+cos(x))/ x^2

9. anonymous

***(3+cos(x))/ x^3 sorry

10. anonymous

Ok nevermind, i got it, thanks!

11. anonymous

Yep. the point is to find a function that is infinitely larger and if that function converges then the smaller one converges. However if the function ( f(x) for example) you test is divergent then f(x) >= g(x) >= 0 then g(x) can't be said is divergent. but if the function ( g(x) in this case) you test for is smaller and divergent then f(x) is divergent.

12. anonymous

so what if the interval was not to infinity?

13. anonymous

if the function is convergent on the interval and larger (meaning on discontinuities), the function is only to be said convergent on that interval.

14. anonymous

No*