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?