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

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 $x+x+\cdots=2$.

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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.