Comparison Test: converges or diverges?

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SUMMARY

The discussion focuses on determining the convergence of the improper integral from 0 to infinity of the function (e^x)/[(e^(2x))+4]. The Comparison Test is employed to analyze the convergence, where the user initially compares the function to g(x) = e^x. However, since the integral of e^x diverges, this comparison is ineffective. The conclusion is that while g(x) diverges, the original function f(x) converges, indicating the need for a different comparison function that effectively demonstrates convergence.

PREREQUISITES
  • Understanding of improper integrals and their convergence
  • Familiarity with the Comparison Test in calculus
  • Knowledge of exponential functions and their properties
  • Basic skills in evaluating limits as they approach infinity
NEXT STEPS
  • Research alternative comparison functions for the Comparison Test
  • Learn about the properties of integrals involving exponential functions
  • Study the implications of constants in the denominator on convergence
  • Explore advanced techniques for evaluating improper integrals
USEFUL FOR

Students in calculus courses, educators teaching integral calculus, and anyone seeking to deepen their understanding of convergence tests for improper integrals.

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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?
 

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