-2
$\begingroup$

I have two issues related to finite sum of infinite series,
1) How you would to describe 2 when you talk about the infinite geometric series 1+ 1/2 + 1/4 + 1/8 + .....
2) How you would compare using partial sums in following cases,
1-1+1-1+...... = 1 -(1-1+1-1+...) which gives 1-1+1-1+... = 1/2 With the method we apply to express recurring decimals as fractions ( ratio between two integers)

  1. Which way is correct , 2 is the sum of infinite number of terms or 2 is the limit of sum of n number of terms as n tends to infinity. In exam papers I have seen students are asked to find sum of infinite number of terms.
    2)As we know 1/2 can not be accepted as the sum of 1-1+1-.... then what about applying the same method in recurring decimals, there we take x as the recurring part and subtract two equations to remove x to get rational form of the number. Where x is sum of infinite series.
$\endgroup$
17
  • 2
    $\begingroup$ After students take a good calculus course, and learn proper definitions related to infinite series, then they could answer such questions. But before that time, when all they have seen are "popularized" expositions, these questions should not be placed on exam papers. $\endgroup$ Oct 28, 2022 at 8:37
  • 3
    $\begingroup$ @JanakaRodrigo: There are numerous books on calculus and/or analysis where infinite series are explained in detail. I suggest that you check a few of them, and then choose those explanations from those books that you find the most suitable for your students. $\endgroup$ Oct 28, 2022 at 9:56
  • 3
    $\begingroup$ In the second situation, both objects $1-1+1-\dots$ and $-1+1-1+\dots$ do not exist (since for each of them, the partial sums do not converge). So there is also nothing to compare because it does not make sense to compare non-existing objects. $\endgroup$ Oct 28, 2022 at 13:30
  • 1
    $\begingroup$ This is a math question, not a math education question. $\endgroup$
    – Sue VanHattum
    Oct 28, 2022 at 15:44
  • 1
    $\begingroup$ And the answers to these questions are found in most if not all textbooks used for this part of the calculus sequence. $\endgroup$
    – Sue VanHattum
    Oct 28, 2022 at 19:04

1 Answer 1

5
$\begingroup$

"Infinite sum" is in common use, so it should be acceptable to say that $2$ is the value of the infinite sum $1+\frac{1}{2}+\frac{1}{4}+\ldots$. But students need to be very clear that the value of an infinite sum is defined to be the limit of the partial sums, assuming that limit exists, and that if the limit doesn't exist, the infinite sum is undefined. They should also be well aware that operations that would not change the value of a finite sum, such as rearranging terms, can change the value of an infinite sum unless it is absolutely convergent. They should be aware of the Riemann rearrangement theorem, which says that the terms of a conditionally convergent series can be arranged to give any value whatsoever, including $\infty$ and $-\infty$.

You ask to compare the calculation $$ x=1-1+1-1+1-\ldots=1-(1-1+1-1+\ldots)=1-x, $$ which leads to $x=1-x$ or $x=\frac{1}{2}$, and calculations like $$ x=0.3333\ldots=-3+3.3333\ldots=-3+10x, $$ which leads to $x=-3+10x$ or $x=\frac{1}{3}$.

The issue here is that the first sum is divergent since the partial sums do not have a limit, making the entire calculation illegitimate (What is $x$ here?) In contrast, the second sum is convergent. It is only because of the convergence that we can meaningfully write $x=0.3333\ldots$. Without convergence $x$ would be undefined. Once we have convergence we can prove the other needed statement, namely that $3.3333\ldots=10x$. All infinite decimal numbers can be seen to be convergent by, for example, using the comparison test with the convergent geometric series $0.999\ldots$.

$\endgroup$
2
  • $\begingroup$ Thanks, well explained. $\endgroup$ Oct 28, 2022 at 15:09
  • $\begingroup$ Just as a note, most of the issues with divergent series go away if you use the more specific hyperreal number system rather than just the reals. For instance, the Riemann Rearrangement Theorem is no longer valid when using hyperreal numbers. The rearrangement theorem actually depends on an ambiguity in the conception of infinite numbers from the perspective of the real line. Hyperreal numbers remove the ambiguity and therefore the theorem. See my paper "Hyperreal Numbers for Infinite Divergent Series" arxiv.org/abs/1804.11342 $\endgroup$
    – johnnyb
    Nov 9, 2022 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.