Some of this blog post is kind of embarrassing to write, it regards an idea that I have worked with for years and I feel I should have had a better sense of it than I do. I’ll start with a conversation I had with my wizards in BC Calculus recently.
We are discussing convergence tests for series and one of the standard tests is the integral test. As Stewart presents it, at least how mrdardy presents Stewart’s presentation, we discuss an infinite series and if the integral of the function has a finite answer, then the series itself converges. Formally, what my students were told is this:
Suppose f is a continuous, positive, decreasing function on the interval [1,∞) and let the sequence a(n) = f(n). Then the infinite series ∑a(n) is convergent if and only if the improper infinite integral ∫f(x)dx is convergent. If the integral is convergent, then the series is convergent. If the integral is divergent, then the series is divergent.
Pretty clean and clear for a calculus statement, right? We had a nice discussion about the difference between looking at the integral which is a continuous summation versus the value of the series which only sums the natural number inputs. A question arose about how to compare the value of the integral with the value of the series. We decided, I really leaned them in this direction by thinking out loud, that the integral sum would be greater than the sum of the series since the integral also added “all that stuff between natural number inputs”. We all seemed happy with that conclusion until yesterday. My LaTex skills are lacking, so I will try this as best I can without that tool. Apologies in advance.
The series we were looking at was the infinite series of 4/(n^2+n). They quickly recognized that this could be rewritten using the method of partial fractions. We rewrote it and, by the magic of the telescoping series, we saw that the sum was 4. All good, right? Well, someone asked me to remind them how we could use the integral test to confirm convergence. We integrated the function using partial fractions and arrived at a sum of 4 ln2. Nice, the integral converged as well. Not so nice, the integral is less than 4. This violated what felt like a nice conceptualization that we had talked through a few days earlier. I sent out a quick call to twitter about why the integral test, that I had convinced myself of providing a ceiling for the sum approximation, would do this. I got a gentle reply from @dandersod reminding me that the integral is the floor for the approximation. I am pretty sure that I have known this at some point, but had constructed such a compelling Riemann approximation argument with my students that I was stumped. I dismissed them for lunch.
One of my advisees dropped by for a quick chat. He is one of our three Differential Equations students this year. He took a look at the problem and I tried to convince him of my mistaken impression that the integral gathers up all this other area and should be an overestimate. He paused, thinking about the problem and then had a great ‘A-ha’ moment. He said, isn’t the series just the left hand Riemann approximation? Since the function is decreasing (otherwise, it wouldn’t converge!) this left-hand approximation will necessarily be an overestimate. Nice, clear reasoning. The kind of reasoning I should have had at my command when talking with my class a week or so ago. Sigh.
My Calculus kiddos are taking a test today. I get to apologize AND praise one of their colleagues on Wednesday when we revisit this ‘mystery’.