# Explain non-convergent sums to a bright high schooler.

I know the higher-math answer to this question, but I'm asking how on Earth to explain it to a bright high school student. Here's the question, paraphrased:

“I can see that $\left(1+x+x^{2}+x^{3}+x^{4}+....\right)\left(1-x\right)=1$ because the terms with non-zero powers of $x$ all cancel out. But if I plug in $x\rightarrow2$, the left factor is $(1+2+4+8+….)$ and the right factor is $-1$. So how can positive times negative equal positive? How can infinite times finite equal finite? This makes no sense! Math doesn't work!”

When I attempt to explain non-convergence or limited domains of convergence, the sharper students react with something like, “What the hell, math just decides to not work sometimes??? So how do we know when we're supposed to disregard logical conclusions??? I no longer trust adults!!!”

Any advice for a stumped tutor? “You’ll understand when you’re older" and "How can we have a unified country if you don't believe the things we tell you?" are not acceptable responses.

• Comments are not for extended discussion; this conversation has been moved to chat. – Jyrki Lahtonen Jun 25 '17 at 12:38
• Please see the matheducators.stackexchange – amWhy Jun 25 '17 at 17:35
• @amWhy There is an education tag on this SE site, and there are many questions here about math education. – Jerry Guern Jun 25 '17 at 21:19
• I suggested what I think is the better location to ask your question, Jerry. It was nothing personal, nor a reflection or demeaning of the quality of your question, but math educators, abundant at the link I give above, are in a better position to discuss student interactions, compare notes, on this issue with other educators. – amWhy Jun 25 '17 at 21:24
• A lot of the participants here are grad students, undergrad, pre-college, some professors, some professors retired from teaching; I think there are fewer people here, with pedagogical knowledge and experience as there are at matheducators.se But, obviously, you want an answer (12, 13, 24, 15) here. So be it. You asked for honest feedback, and I tried to give it; if you want to sulk and insist this question belongs here, don't blame me for how you choose to respond to the limited answers given here. – amWhy Jun 25 '17 at 21:33

I can see that $\left(1+x+x^{2}+x^{3}+x^{4}+....\right)\left(1-x\right)=1$ because the terms with non-zero powers of x all cancel out

This only appears to be the case for the infinite sum. For the finite sum you can see that

$(1+x+...+x^n)(1-x)=1-x^{n+1}$

I think you must explain to your student that the only sensible way to look at an infinite sum is to consider it as a limit of finite sums. You can plainly see that the $x^{n+1}$ term only vanishes for $|x|<1$ so the original statement is only valid for $|x|<1$

• A thousand times this. You can not "see" that that equality is true period. Do the stuff properly or not at all. Look at the limits and don't fill students heads with intuition and similar it's bad enough there is all the -1/12 stuff floating around confusing everyone. – DRF Jun 23 '17 at 6:27
• Yes. The correct “higher math answer” is indeed also the correct high school answer. – Carsten S Jun 23 '17 at 8:06
• @DRF I would definitely agree that intuition can't be a substitute for rigor, but "don't fill students heads with intuition" sounds far too much like "don't even bother trying to cultivate mathematical intuition," which is just...completely anathema to me. – Kyle Strand Jun 23 '17 at 21:54
• @KyleStrand It's unfortunate that in mathematics we only have the one word, "intuition", to cover a wide variety of things, some of which are good and some of which are very bad. In this case, the "intuition" that this identity should be true because "the $x$'s cancel out is not intuition", it's just habit. I'm sure the person you're replying to wouldn't extend the definition of "intuition" in their statement to cover anything you would find objectionable. – Jack M Jun 23 '17 at 22:35
• @JackM I agree that semantics is often frustratingly vague and imprecise! I'm not entirely sure I agree that this is the point of disagreement here, though--I think that the intuition that the identity "seems" true is valuable, because that's the starting point for understanding when it is and when it is not valid. – Kyle Strand Jun 23 '17 at 23:03

The student says "positive times negative is negative". What I would say to that is that you cannot multiply adjectives. What is true (and an interesting discussion in itself) is that

the product of a positive number times a negative number is a negative number.

Next I would ask the student which number is "$1+2+4+8+\cdots$" supposed to be. Followed by how on earth he dares to assume that adding infinitely many numbers actually makes sense. And then explain that, in some situations, one can make sense of an infinite sum; and that there is no reason to expect all rules for finite sums to hold for infinite sums.

In his manipulation, the student is tacitally assuming that $$(1+x+x^2+\cdots)-(x+x^2+x^3+\cdots)=1+(x-x)+(x^2-x^2)+\cdots.$$ With that "logic" I would consider $x=1$ in his equality (that leads to $0=1$) and show that one could do $$(1+1+\cdots)-(1+1+\cdots)=1+(1-1)+(1-1)+\cdots=1$$ or $$(1+1+\cdots)-(1+1+\cdots)=(1-1)+(1-1)+\cdots=0$$ or $$(1+1+\cdots)-(1+1+\cdots)=27+(1-1)+(1-1)+\cdots=27.$$

Finally, as for not "trusting adults", there may be good reasons for that, but math is not one of them.

• I suspect you're not a fluent English speaker. When the adjective is the essential part and the noun is obvious from context, it's common in English to use just the adjective as a shorthand for an entire article-adjective-noun phrase. "Slow and steady wins the race." "Smart beats lucky every time." – Jerry Guern Jun 24 '17 at 14:41
• It depends on your definition of "fluent". I'm not a native speaker, if that's what you mean; still, adjectives (and the use you mention) exist in a few languages other than English. In any case, you are missing the point: for the student in this question, the "adjective-noun" substitution was allowing him to apply the rule to an object that was not a number, without even thinking about it. – Martin Argerami Jun 24 '17 at 23:49
• Your point is valid. You want to outline to the student that infinity is not a number. This is the reason why you cannot apply the 'rules' that you would ordinarily expect. – Jay Jun 25 '17 at 2:32

The Series at Hand

As I mentioned in my comment you could simply show them the original identity behind this,

$$(1+x+x^2+x^3+x^4+\dots) = \frac{1}{1-x} \quad \text{ when } |x|<1$$

I think to most who understand addition and powers, it should seem fairly intuitive that this series doesn't converge for $x \ge 1$ or $x \le -1$ (or in a middle schoolers words "gets bigger forever"). For the student who remarks "Math decides not work," simply explain that this is math working. If something like $1 + 2 + 4 + 8 + 16 + \dots$ had a finite value, then things modeled with exponential growth, like bacteria and decay, would limit themselves. They would stop and approach a value, which is not what we observe in the world.

Break their Presuppositions

As described in this comment, it is important for a mathematician to "retune their intuition to match the truth." Show them several ("less mathy") examples of how our naïve intuition breaks down at infinity. These could be the infinite hotel paradox, Zeno's paradox, Gabriel's horn, this $\pi=4$ question, Banach-Tarski (might be kind of difficult to explain), etc...

Keep in mind this only leaves more questions and does not answer much, which can be good to create a drive to learn more.

• i may also suggest mentioning that these series and regularization is at the current forefront of mathematics. such as the zeta function – shai horowitz Jun 22 '17 at 22:17
• Thank you for the suggestion. The three examples you listed are simply breakdowns of human intuition, e.g., we might naively expect a finite volume to have finite surface area but we'd be demonstrably wrong, case closed. Expecting a sum of positive numbers to stay positive is a more defensible expectation, not as easy to dismiss as mis-intuition. – Jerry Guern Jun 22 '17 at 22:34
• @JerryGuern However, when $|x|<1$, the terms get smaller and approach zero. This, I mean, intuitively has a finite value (ofc I wouldn't know how to explain the diverging harmonic well). As for the positive-negative dilemma, this shouldn't be an issue if you show them $x \to 2$ is undefined (the first half of my answer). – Dando18 Jun 22 '17 at 22:42
• Great answer, but be careful with the "middle schoolers words": "gets bigger forever" - I understand what you meant, but this might confuse students, since there are strictly increasing sequences that still converge. – Pedro A Jun 23 '17 at 13:37
• @Hamsteriffic yeah I see what you mean, I meant that's what the middle schooler would think. – Dando18 Jun 23 '17 at 14:14

I think the simplest answer is that the laws of arithmetic do not work, and were never intended to work, for infinities.

When said student was younger, he was taught, for example, that $x-x=0$. This is true for real numbers, but for infinities the operation is not defined ($\infty-\infty$ would be the unique number satisfying $\infty+x=\infty$, but this number is not unique).

He can't just assume that he can go and manipulate expressions diverging to infinity nilly-willy without formal justification.

If he does build a formally justified framework, one of two things will hold:

1. He will find that what he tried to do is not a permissible manipulation.
2. The end result is correct, because he is working with 2-adics or something similar.

One should start at the definition of an infinite series to see what is going wrong. We have

$$\sum_{n=0}^\infty a_n=\lim_{N\to\infty}\sum_{n=0}^Na_n$$

In this case, we have

$$\sum_{n=0}^\infty x^n=\lim_{N\to\infty}\sum_{n=0}^Nx^n$$

Now, I will draw attention to the partial sums: As your students say, there is some cancellation:

$$(1-x)\sum_{n=0}^Nx^n=1-x^{N+1}$$

However, in the partial sums, not everything cancels as they claim. We have an extra $x^{N+1}$ here. Note that when $|x|<1$, the remainder goes to zero, hence,

$$|x|<1\implies\sum_{n=0}^\infty x^n=\frac1{1-x}$$

On the other hand, if $|x|\ge1$, then

$$x^{N+1}\not\to0\implies\sum_{n=0}^\infty x^n\ne\frac1{1-x}$$

"Then what have I done wrong?"

To which you should respond:

"You cannot cancel infinitely many terms, only finitely many terms. The 'last terms' of the infinite series must tend to zero, as you cannot cancel them yourself."

The important lesson here is that one should draw careful attention to the partial sums when doing any arithmetic with infinite series, even if it does not appear you've done any wrong. A particular example is the Riemann series theorem which shows that even if some algebraic manipulation of a convergent series yields a finite result, the result may not be correct.

On the other hand, they may be perplexed to wonder:

"Does $-1$ hold any meaning to $1+2+4+8+16+\dots$?"

To which you may carefully respond:

"Yes."

Borrowing from Jyrki Lahtonen's comment,

"Actually $1+2+4+8+16+\dots$ is equal to $−1$. But only in the field of $2$-adic numbers (where the series also converges). Actually, it may be easier to convince a computer programmer about this. If you could use infinitely many bits to represent integers, then $−1=\dots111111_2$. Just like in the twos complements that they are familiar with. Mathematically, a key difference between $2$-adic numbers and the usual numbers is that the former don't have an order relation. From a $2$-adic number you cannot say whether it is positive or negative, so the problem you pointed out does not arise."

There happens to be an alternative explanation though, as I will give. If we define:

$$f(x)=\sum_{n=0}^\infty x^n;\quad |x|<1$$

And $f:\mathbb C\setminus \{1\}\mapsto\mathbb C\setminus0$ and $f\in C^\omega$ on its domain.

Then,

$$f(2)=-1$$

What $f\in C^\omega$ means is that $f$ is an analytic function, which can be uniquely defined through analytic continuation. The process is usually tedious, but it is not so bad here. More or less, if your high schoolers know of differentiation or Taylor's theorem, then,

$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n$$

For any $a\in\mathbb C\setminus1$ and $|x-a|<R$ where $R$ is the radius of convergence. For example,

$$f(-1)=\sum_{n=0}^\infty\frac{f^{(n)}(-0.5)}{n!}(-0.5)^n=\frac12$$

And this occurs even though $f(x)$ is not directly defined at $x=-1$. Note that this is a well-defined convergent series and since $f(x)$ is defined for $|x|<1$, then $f^{(n)}(-0.5)$ is defined, and with some work it can be shown that the above indeed reduces to a geometric series. Indeed, when working with this, one finds that:

$$f(x)=\frac1{1-x}$$

And likewise, we can deduce that

$$f(2)=-1$$

However, it is not correct to state that

$$f(x)=\sum_{n=0}^\infty x^n\implies f(2)=-1$$

The important component is the requirement that $f$ is an analytic function.

It is also very important to note that $f(x)$ existing for $|x|\ge1$ does not mean that $\sum_{n=0}^\infty x^n$ converges for $|x|\ge1$ nor should one associate $f(2)$ to $\sum_{n=0}^\infty2^n$.

As a side note, one can extend $f(x)$ to $x<1$ by repeatedly applying Taylor expansions at $x=-t,t>0$, but to extend $f(x)$ to $x>1$, you will have to Taylor expand in the complex plane.

In the same fun manner, we may consider a well known function:

$$\zeta(x)=\sum_{n=1}^\infty\frac1{n^x},x>1$$

If we consider $\zeta(x)$ to be analytic, it is possible to make sense of $\zeta(0),\zeta(-1),\dots$, which have well known values of $-\frac12,-\frac1{12},\dots$, but it is, again, important not to confuse this with the divergent series:

$$\sum_{n=1}^\infty1\ne-\frac12,\quad\sum_{n=1}^\infty n\ne-\frac1{12}$$

If you are looking for rather elementary approaches to deriving $\zeta(-k)$, I recommend starting with:

$$\eta(x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^x},\quad x>0\\\eta(x)\text{ is analytically extended to all }x\in\mathbb C$$

And then when $x>1$, show that

$$\zeta(x)-\eta(x)=2^{1-x}\zeta(x),\quad x>1$$

"Watch those partial sums!"

The second line is a functional equation that may be used to relate $\zeta(x)$ and $\eta(x)$ for any $x\in\mathbb C\setminus \{1\}$, so we may use it to see that

$$\zeta(x)=\frac{\eta(x)}{1-2^{1-x}}$$

And if you show that $g(x)$ is an analytic function, where

$$g(x)=\lim_{t\to-1^+}\sum_{n=1}^\infty\frac{t^{n+1}}{n^x}$$

and that $g(x)$ is both defined for all $x\in\mathbb C$ and equal to the $\eta(x)$ function, then you may calculate values of the Riemann zeta function. For example,

\begin{align}\zeta(0)&=\frac{\eta(0)}{1-2^{1-0}}\\&=-\lim_{t\to-1^+}\sum_{n=1}^\infty t^{n+1}\\&=-\lim_{t\to-1^+}\frac{t^2}{1-t}\\&=-\frac12\end{align}

So I assure you, the math does work, but it's not as simple as a telescoping algebraic manipulation.

I think the existing answers are pretty unconvincing or hard to read, especially in light of the whole -1/12 business. Here's what I think is a better approach:

Point out to him that the "reason" we have $$1 + x + x^2 + \ldots = \frac{1}{1 - x}$$ is NOT that the terms "cancel". Rather, it's that if the left side had a finite sum $S$ $$S = 1 + x + x^2 + \ldots$$ then we can prove that $S$ must satisfy the following: \begin{align*} S x &= x + x^2 + x^3 + \ldots \\ 1 + S x &= 1 + x + x^2 + \ldots \\ 1 + S x &= S && \text{ (true by our assumption)} \\ (S - S x) &= 1 \\ S (1 - x) &= 1 \\ S &= \frac{1}{1-x} \end{align*}

However—and this is key—we have merely proved a CONSTRAINT on such a sum $S$.
We did NOT prove that $S$ exists at all! We simply assumed it and derived a property that $S$ satisfies.

To prove that the the two sides are equal, then, we must first define what it means for a finite series to be "equal" to an infinite series, since, until now, we've only had a meaning for equality for finite expressions.

Of course, a very reasonable definition for this (which I assume they have been taught) is that the partial sums of the series converge to a limit $S$ (using the epsilon/delta definition of a limit).
And of course, once we use that definition, then all you need to do is show that the partial sums do (or don't) converge to $S$, and only after doing that can we declare $S$ to be equal to $1 + x + x^2 + \ldots$.

[Optional] However, if you change your definition of equality of a finite and an infinite series, you can obtain a different result. For example, if you simply define the sum $S$ of an infinite series to be "whatever number $S$ that satisfies the same algebraic expressions as the original series", then the student's statement will be correct. You should point out that this is not the definition of equality they have been taught, however, as there are many pitfalls to this approach, and it is useless (if not outright harmful) for most practical purposes, including the the fact that—as he rightly pointed out—such a definition would violate many intuitions.

Nonetheless, I think if you explain the above clearly, he should be able that clearly see that the math he has already learned works either way, and that there is no contradiction whatsoever.
The only contradiction comes up if he decides to define equality differently than you have.
Which is fine, because nobody guaranteed different definitions would lead to the same conclusions!
(P.S.: This is also a good way to teach why definitions matter.)

Take the series $1+(-1)+1+(-1)+\cdots$. Your student will probably think the sum is 0 because you can cancel out the $1$'s with the $(-1)$'s. That's because they don't know the definition of a series: a series is the limit of the sequence of its partial sums.

The sequence of partial sums in the series $\left(1+x+x^{2}+x^{3}+x^{4}+\cdots\right)\left(1-x\right)$ doesn't converge for $x=2$. So the sum of the series $\left(1+x+x^{2}+x^{3}+x^{4}+\cdots\right)\left(1-x\right)=1$ is not defined.

• See in my Question the response I received when I tried to explain limited domains of convergence. – Jerry Guern Jun 22 '17 at 22:19

The problem in my opinion is that the following statement

I can see that $\left(1+x+x^{2}+x^{3}+x^{4}+....\right)\left(1-x\right)=1$ because the terms with non-zero powers of $x$ all cancel out.

is not correct. The conclusion is "kind of" correct, but the reasoning is definitively not. To solve the problem, we must simply correct the reasoning. So let's go back to some basics and calmly apply them.

First, what is an infinite sum ? It is defined as

$$\sum_{k=0}^\infty a_k := \lim_{N \rightarrow \infty} \sum_{k=0}^N a_k.$$

It is not about writing lots of plus signs with dots at the end, it is really the limit of something. Then let's put our example inside, it yields

$$(1 - x) \sum_{k=0}^\infty x^k= \lim_{N \rightarrow \infty} (1-x) \sum_{k=0}^N x^k = \lim_{N \rightarrow \infty} \left[\sum_{k=0}^N x^k - \sum_{k=0}^N x^{k+1} \right].$$

The statement I quoted from your question basically comes down to assert that

$$\lim_{N \rightarrow \infty} \left[\sum_{k=0}^N x^k - \sum_{k=0}^N x^{k+1} \right] = \lim_{N \rightarrow \infty} \sum_{k=0}^N x^k -\lim_{N \rightarrow \infty} \sum_{k=0}^N x^{k+1} \\ = \sum_{k=0}^\infty x^k - \sum_{k=0}^\infty x^{k+1} = 1 + \sum_{k=0}^\infty x^{k+1} - \sum_{k=0}^\infty x^{k+1} = 1.$$

The second line is perfectly fine, but the first equality only holds only if one of the terms converges [*], which is not the case in general. Therefore the reasoning is flawed, and you have to do it in an other way if you want a correct result[**]. The simplest is to do the labelling before performing the limit

$$\lim_{N \rightarrow \infty} \left[\sum_{k=0}^N x^k - \sum_{k=0}^N x^{k+1} \right] = \lim_{N \rightarrow \infty} \left[1 + \sum_{k=0}^N x^{k+1} - \sum_{k=0}^N x^{k+1} - x^{N+1}\right] \\ =\lim_{N \rightarrow \infty} \left[1 - x^{n+1} \right] = 1 - \lim_{N \rightarrow \infty} x^{N+1}.$$

The domain of convergence is thus explicit.

The conclusion is that if you follow the rules, mathematics gives you correct results. If you do not, you can not expect anything sensible. You even have to be careful to be sure to follow the rules correctly.

For me this problem is of the broader class of problems that people learning mathematics encounter that I call "Let's use that unproven short-cut, what could go wrong ?".

[*] To have the limit of a sum to converge you really need both terms to converge, but if you allow to write

$$x \pm \infty = \pm \infty, \qquad \text{for } \vert x \vert < \infty,$$

then the statement is correct.

[**] If you do not care about having correct result, the problem is not the mathematics.

As I see this is not a popular opinion but I'd say that it is better to avoid the notion of convergence in this explanation. Or talk about convergence a bit later.

The proper answer is that math always works but as the "proof" was about $x$ and not about real numbers, the consequences of plugging $2$ into this "picture of $x$" manipulation should not come as a surprise.

Because the fact the high schooler noticed is obviously true it must hold in some formal sense. And indeed it does: this formal power series identity can be justified as an infinite column of identities, each in its own ring $\dfrac{\mathbb{Q}[[x]]}{(x^n)}$ (or not over $\mathbb{Q}$ ... whatever).

You definitely should not tell all the theory to one high schooler at once but if you replace "formal power series" with "some picture manipulation" it would be a nice explanation IMO.

After that, you can talk about summation methods but it it is not compulsory :)

• I think I was having a thought along this line -- when you do these manipulations, they must be valid in some sense. But could you explain what you mean about your "infinite column of identities" and why it's $\dfrac{\mathbb{Q}[[x]]}{(x^n)}$? – Luke Jun 24 '17 at 18:03

It is helpful to explain the students the concept of sequence of numbers, in particular, the arithmetic and geometric sequences. Next one can distinguish an increasing and decreasing geometric sequences and discuss the sum of terms of the geometric sequence. Next one can get the students to understand the sum of all terms of descending (ascending) geometric series is finite (infinite). And then one can discuss the switch from the sum of $n$ terms to the sum of all terms of geometric series. That is: $$1+x+x^2+...+x^n=\frac{1\cdot(1-x^{n+1})}{1-x}\Rightarrow 1+x+...+x^n+...=\frac{1}{1-x}, |x|<1.$$

As a classical example for the sum of terms of decreasing geometric series, one can consider consecutively halving the rectangle of area $1$ and adding the areas of smaller rectangles to total $1$.

In conclusion, the mathematics is constructive, that is, the more advanced notions are based on the simpler notions.