Dig-In: Remainders and the Integral Test

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We investigate how the ideas of the Integral Test apply to remainders.

We have split up our infinite sum $S$ into two important pieces: a finite piece $S_n$ , which is our estimate, and an infinite piece $R_n$ , which is our error or remainder.

We will focus mostly on the remainder, since the estimate is relatively easy to obtain, especially with the use of a computer. Even though the remainder is an infinite sum, generally we will be concerned with ensuring that the remainder is small – using whatever definition of “small” fits the needs of our problem or situation.

Let’s begin with a series $\sum a_k$ and assume that this series converges, and that we would like to estimate its sum. Let $f(x)$ be the function associated with the sequence $a_k$ , so that $f(k) = a_k$ . Assume this function $f$ is eventually continuous, positive, and decreasing, so that the Integral Test applies.

Since the Integral Test applies, we know that the integral $\int _N^{\infty } f(x) \, dx$ converges to some finite value $L$ , but that $L$ is not the sum of the series $\sum a_k$ . While explaining why the Integral Test makes sense, we imagined the terms of our sequence to be a particular Riemann Sum for the integral in question.

We can leverage this same picture into helping us estimate the remainder $R_n = \sum _{k=n+1}^\infty a_k$ , by imagining that we are trying to add up only those rectangles corresponding to $R_n$ . Remember that we should begin with $a_{n+1}$ , or the box whose width is $1$ and whose height is $f(n+1)$ .

Notice that all of the rectangles representing the terms of $R_n$ are completely below the graph of $f(x)$ . In symbols, $R_n = \sum _{k=n+1}^\infty a_k < \int _n^\infty f(x) \, dx.$ The error is bounded above by the integral. Assuming we can evaluate the integral, we can give an upper bound for the error $R_n$ .

Consider the series $\sum _{k=1}^\infty \frac {1}{k^2}$ . Give an upper bound for the error involved with using $S_4$ in place of $S$ .

The question is asking us to overestimate the value of $R_4$ . Since $f(x) = \frac {1}{x^2}$ is continuous, positive, and decreasing on $[1, \infty )$ , let’s use the Integral Test to make our overestimate.

We have just seen that $R_4 < \int _{\answer [given]{4}}^{\infty } \frac {1}{x^2}\, dx.$

We can evaluate the integral above. $\int _{4}^{\infty } \frac {1}{x^2} = \lim _{b \to \infty } \int _4^b \frac {1}{x^2}\, dx = \lim _{b \to \infty } \left [ \frac {1}{\answer [given]{-1}}x^{\answer [given]{-1}} \right ]_4^{\infty } = \answer [given]{0.25}$

Since $R_4$ is less than the value of this integral, we find that $R_4 < \answer [given]{0.25}$ .

In the event that our application requires us to work with an overestimate for the sum $S$ , we can combine our information about $R_n$ with our estimate $S_n$ to get an upper bound for $S$ . In other words, $S = S_n + R_n < S_n + \int _{n}^{\infty } f(x)\, dx.$

Consider again $\sum _{k=1}^\infty \frac {1}{k^2}$ . Using $n=4$ and the information we found in the previous example, what is an upper bound for the value of this series? $S < S_n + \int _4^{\infty } \frac {1}{x^2} \, dx = \answer [given]{1 + \frac {1}{4} + \frac {1}{9} + \frac {1}{16}} + \answer [given]{0.25} = \answer [given]{1+\frac {1}{4} + \frac {1}{9} + \frac {1}{16} + 0.25}$

Recall that there are two types of questions we can ask about remainders.

(a): What is the error involved with using a specified number of terms?
(b): How many terms of a series (what is the value of $n$ ) should we use to obtain a desired precision?

We’ve covered the first of these questions; let’s turn our attention to the second.

How many terms of the series $\sum _{k=1}^{\infty }\frac {1}{k^2}$ do we need to add in order to guarantee that the estimate $S_n$ is within $0.01$ of the actual sum $S$ ?

This question is asking us to find a value of $n$ which will guarantee that $S - S_n < 0.01$ . Since the function $f(x) = \frac {1}{x^2}$ is continuous, positive, and decreasing on $[1, \infty )$ , let’s use the Integral Test to find our answer.

The Integral Test gives us that for any value of $n$ , $R_n < \int _{\answer [given]{n}}^{\infty } \frac {1}{x^2}\, dx = \frac {1}{n}.$ Since the remainder is less than the value of the integral, if we make the value of the integral less than $0.01$ , then the remainder will also be less than this value. In symbols, if for a given value of $N$ $R_N < \int _N^{\infty } f(x)\, dx < 0.01,$ then $R_N < 0.01$ for that same value $N$ . In our case, if we want $R_n < \int _n^{\infty } \frac {1}{x^2}\, dx = \answer [given]{\frac {1}{n}} < 0.01$ by solving the second inequality for $n$ , we can see that this is true for $n > \answer [given]{100}$ . In other words, if $n > \answer [given]{100}$ , then $R_n < 0.01$ , as requested.

Notice that in the previous example, any larger value of $n$ will also do. So, as long as we add up at least a hundred terms, the value of $S_n$ will be within $0.01$ of the sum $S$ . Since we know in this case that $\sum _{k=1}^\infty \frac {1}{k^2} = \frac {\pi ^2}{6}$ , we could check this ourselves!

Finally, it’s worth noting that the Integral Test can actually take us quite a bit further than we’ve gone here. You might have already noticed that, with the proper setup, the value of the integral can provide an estimate for the entire sum $S$ . Whether this estimate is good enough depends, of course, on your situation. Furthermore, if we had been clever about how we had organized our rectangles corresponding to the sequence $a_k$ and a Riemann Sum for the integral, we could also come up with a lower bound for the error. Creating such a lower bound is not as frequently useful as creating an upper bound – but again, this depends entirely on your application!