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.
![[Picture]](digInRemaindersAndTheIntegralTest.online-02a2d2d6da1a9561f23b9609d3a868ba.svg)
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.
![[Picture]](digInRemaindersAndTheIntegralTest.online-ced822a46e9f74edfbe22a34bad5e524.svg)
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 \(f(n+1)\).
![[Picture]](digInRemaindersAndTheIntegralTest.online-06fd9df40abfa92f8b6bf6d019fb5b6e.svg)
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!