The Unapologetic Mathematician

Mathematics for the interested outsider

Absolute Convergence

Let’s apply one of the tests from last time. Let \alpha be a nondecreasing integrator on the ray \left[a,\infty\right), and f be any function integrable with respect to \alpha through the whole ray. Then if the improper integral \int_a^\infty|f|d\alpha converges, then so does \int_a^\infty f\,d\alpha.

To see this, notice that -|f(x)|\leq f(x)\leq|f(x)|, and so 0\leq|f(x)|+f(x)\leq2|f(x)|. Then since \int_a^\infty2|f|\,d\alpha converges we see that \int_a^\infty|f|+f\,d\alpha converges. Subtracting off the integral of |f| we get our result. (Technically to do this, we need to extend the linearity properties of Riemann-Stieltjes integrals to improper integrals, but this is straightforward).

When the integral of |f| converges like this, we say that the integral of f is “absolutely convergent”. The above theorem shows us that absolute convergence implies convergence, but it doesn’t necessarily hold the other way around. If the integral of f converges, but that of |f| doesn’t, we say that the former is “conditionally convergent”.

April 22, 2008 - Posted by John Armstrong | Analysis, Calculus | | 2 Comments

2 Comments »

  1. [...] also can import the notion of absolute convergence. We say that a series is absolutely convergent if the series is convergent (which implies that [...]

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  2. [...] Ratio and Root Tests Now I want to bring out with two tests that will tell us about absolute convergence or (unconditional) divergence of an infinite series . As such they’ll tell us nothing about [...]

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