Thread: [SOLVED] Integral convergence and divergence:How to choose comparison function?

1. [SOLVED] Integral convergence and divergence:How to choose comparison function?

I'm having difficulty understanding direct comparison test and limit comparison test for
determining integral convergence and divergence. For direct comparison test how do you
choose the second function?

Can I arbitrarily choose any function for comparison with the given function Or is their
any criteria for choosing the function? Also for limit comparison test how do you
figure out the second function? Any help will be appreciated.

2. I'll try my best to give some sort of input, however I may be wrong also before I try to help this link can be very good for the concept of using comparison test which is mainly used on improper integrals or rather difficult integral What I did I followed each example they did and found some sort trigger that makes them choose the method or comparative integral they did

Pauls Online Notes : Calculus II - Comparison Test for Improper Integrals

What you usually you want to observe all the terms in the function and find a term that dominates the function such as the example, very simple example

$\frac{1}{x^2+1}$

Now as x approaches positive infinity in this case I am trying to explain the term
$x^2$ reaches positive infinity much faster than the term or value of $+1$ next too it and say the term will dominate the value of the function in this case, so you can say that integral behaves like the function of
$\frac{1}{x^2}$ which we know by the P-Test converges to a value of 0 If, I recall with accuracy

also what I like to do for some examples is take the denominator if it dominates the function and break it down like this, Ill use my example above

$\frac{1}{x^2 + 1}$

As I said the $x^2$ term will dominate the function I usually break it down from the denominator as such

${x^2+1} > {x^2}$

It obvious to see why this statement above holds true, if not just plug in some values, next to make it comparative like the original which was a under a 1

$\frac{1}{x^2+1}$ and $\frac{1}{x^2}$

Also keep in mind the value $\frac{1}{x^2}$ for large value under 1 will approach 0, so in essence the bigger the value under the 1 the closer it is to 0 so then this statement will hold true if we keep in mind the statement I made above

${x^2+1} > {x^2}$

which over becomes

$\frac{1}{x^2+1} < \frac{1} {x^2}$
Can you see why this is true? If so, then it is easy to know by the P Test that $\frac{1}{x^2}$ converges to a value of a 0, or you can work out the limit of the integral of $\frac{1}{x^2}$ to see that it approaches 0 and converges so therefore our originial integral converges

$\frac{1}{x^2+1}$

we also must keep in mind the rule of the comparison test, if you choose some function $g(x)$ greater than the original function $f(x)$ if it converges ONLY then the original converges, if the greater function $g(x)$diverges this doesn't mean the function $f(x) converges$and the other part of the rule says if you choose some function smaller than the original diverges so will the original hope I didn't confuse you and offered some insight

So you typically want to find some dominate term in the function if you can and take a "guess" if the original function f(x) diverges or converges and you confirm this "guess" by using a comparison to the integral

3. ...Interesting. Thank you so much for taking the time to explain this problem so nicely. The