Home > Calculus > Direct Comparison Test for Convergence of an Infinite Series

How do you use the comparison test to test for convergence if #1 / (xcosx) dx# from 0 to #pi/2#?

Answer 1

Emma Aldridge

Compare to #1/x^2#.

For #x# close to #0#, we have #x < cosx# so #1/cosx < 1/x# and #1/(xcosx) < 1/x^2#

#int_0^(pi/2) 1/x^2 dx# converges, so does #int_0^(pi/2) 1/(xcosx) dx#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Charles Anctil

To use the comparison test for the convergence of the integral (\int_0^{\pi/2} \frac{1}{x\cos(x)} , dx), we need to find a function (f(x)) that is greater than or equal to (\frac{1}{x\cos(x)}) for all (x) in the interval ((0, \pi/2]) and whose integral is known to converge or diverge.

Let's consider (f(x) = \frac{1}{x}). For (x) in the interval ((0, \pi/2]), (x) is positive, and (\cos(x)) is also positive, so (x\cos(x) > 0). Hence, (f(x) = \frac{1}{x} \geq \frac{1}{x\cos(x)}) for all (x) in ((0, \pi/2]).

Now, we know that (\int_1^{\infty} \frac{1}{x} , dx) diverges (harmonic series), so by the comparison test, since (\frac{1}{x} \geq \frac{1}{x\cos(x)}) for all (x) in ((0, \pi/2]), and (\int_1^{\infty} \frac{1}{x} , dx) diverges, then (\int_0^{\pi/2} \frac{1}{x\cos(x)} , dx) also diverges.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.