The Unapologetic Mathematician

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”.

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Posted by John Armstrong | Analysis, Calculus

3 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 [...]

Pingback by | April 25, 2008 | Reply

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 [...]

Pingback by | May 5, 2008 | Reply

3. [...] integral, we have analogues of the direct comparison and limit comparison tests, and of the idea of absolute convergence. Each of these is exactly the same as before, replacing limits as approaches with limits as [...]

Pingback by | January 15, 2010 | Reply

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