Infinite summation: $x+x+x+x+... =2$?

One of my favourite little math problems is this

$x^{x^{x^{x^{...}}}}=2$

The solution to it is quite simple. An infinite tower of x's is equal to 2, and above the first x there is still an infinite tower of x's, so the equation can be simplified to

$x^2=2 \Rightarrow x= \sqrt2$

(Note: this only works iff $e^{-e} \leq x \leq e^{\frac{1}{e}}$)

Now, what if instead of an infinite exponentiation it would be an infinite summation, like this:

$x+x+x+x+...=2$

If we try solve it the same way as the exponentiation one: An infinite sum of x's is equal to 2, and after the first x there is still an infinite sum of x's, so the equation can be simplified to

$x+2=2$

From which it follows that $x=0$, but surely it can't be that $0+0+0+0+...=2$

Is this because $0+0+0+0+... = 0 \times \infty$, which is indeterminate? Or what is going on?

• Geez, what's with the downvotes here? Yes, the OP is misunderstanding what he's doing, but he's then doing the right thing and asking about what goes wrong so he can fix his misunderstanding. That's a perfectly good question! Feb 21, 2016 at 11:00
• @HenningMakholm I agree. Sometimes is very hard to understand why somebody downvotes questions or answers. It looks like something randomly done. Perhaps because anonimity? Feb 21, 2016 at 11:38
• What is going on? Exactly the same thing that happens for $x^{x^{x^{x^{\dots}}}}$ if $x\notin[e^{-1},e^{1/e}]$. Feb 21, 2016 at 13:01

What you have is a correct argument that IF $x$ is such that $x+x+\cdots=2$, then $x=0$.

However, this does not necessarily mean that the converse is true: You have no argument that if $x=0$ then $x+x+\cdots=2$.

So your argument, in combination with the easy fact that $0+0+\cdots\ne 2$, shows you that there is NO $x$ such that $x+x+\cdots=2$.

This has nothing to do with being "indeterminate". It is plain and unambiguous that $0+0+\cdots$ is $0$. When we say that $0\times \infty$ is indeterminate, the only thing we mean is that when you have a limit of the form $\lim_x f(x)g(x)$ where $f(x)$ goes to $0$ and $g(x)$ goes to $\infty$, then knowing the limits of $f$ and $g$ does not tell you what the limit of the product will be. In particular "indeterminate" does not mean that $\lim_x f(x) g(x)$ itself is somehow bad or doesn't exist -- only that you need to work harder to find it than just taking the limits of $f$ and $g$ separately.

• So if x+x+x...=2, then x=0, but if x=0, then x+x+x...=/=2? Is it kind of like an extraneous root? Feb 21, 2016 at 11:42
• @user265554: Yes, that's one way to think of it. Because $x+x+\cdots=2$ is always false, implication with that as its premise will be true. A more vivid example would be "every unicorn has seven legs" (true because there are no unicorns), but "a thing with seven legs is not a unicorn" (true, again, because nothing is a unicorn). Feb 21, 2016 at 11:52

An analogy can be like this:

A geometric point as such has no definite length, i.e. its length can be said to be $0$.

So infinite such points should have a length $= 0 \times \infty$

But infinite such points can form a straight line, a circle, a conic or any other geometric curve with a specified length ($\not = 0$).

Hence,

$$0 \times \infty = \text{finite}$$

Can you see how this is linked with your problem?

$x+x+\dots+x=kx$ say, so we can say your equation is as $k\to\infty$ and $$\lim_\limits{k\to\infty}kx=2$$

I would proceed with $x=\lim_\limits{k\to\infty}\dfrac2 k$, so $x=0$.

As said in the other answer your problem is that you can't materialize the infinite sum of $x$ because it does not converge and if it did then indeed $x$ would be equal to 0.

Another thing which is a "paradox" with unconverging series, try to compute the sum of $2^i$. If you materialize this object, you say there exists a $S$ such that $S=\sum_i 2^i$ but then you have $S- 2S =1$ and then $S=-1$ which means that summing positive number can give you a negative one! What's fun though is that there are ways to formalize those calculations and give them precise sense and it's used in fundamental physics! (If you're familiar with bit decomposition on computer, you can also understand as a full 1-bitset is -1 which is true on a computer by design but it's fun to see that it might be more profound!)