Comparison Test: converges or diverges?

theRukus
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Homework Statement


Determine whether or not the improper integral from 0 to infinite of (e^x)/[(e^2x)+4] converges and if it does, find it's definite value.


Homework Equations





The Attempt at a Solution


I missed the lecture on the Comparison Test, so I'm essentially useless.

I assign g(x) = e^x. Let f(x) be the function defined in the question statement. g(x) > f(x) on 0 to infinite, so if g(x) converges then f(x) converges, correct? I then sub in t as the upper limit and evaluate the function as t approaches infinite. This limit cannot be evaluated, so g(x) converges and therefore f(x) converges, right?

Thanks for any help.
 
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I'm not sure I follow what you're saying, but here's the basic idea behind the comparison test: let a(n) be the series whose convergence you are trying to determine. If you can find some series b(n) such that |b(n)| > |a(n)| for all n (or at least sufficiently large n), and b(n) converges, then a(n) converges (think of it graphically - b(n) squeezes a(n) to zero). Similarly, if |b(n)| < |a(n)| for sufficiently large n, and b(n) diverges, then a(n) diverges.

In this case, you've used as your comparison e^x, which as you've pointed out is always bigger than f(x). However,

\int_0^{\infty} e^x\ dx diverges, so that doesn't do you any good. You are correct, however, that f(x) converges. Try to think of another function larger than f(x). As a hint: what does that 4 in the denominator do for you?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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