Explain non-convergent sums to a bright high schooler.

Solution 1:

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 $|x|<1$

Solution 2:

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.

Solution 3:

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.

Solution 4:

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:

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

Solution 5:

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

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

They may then ask:

"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,

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