The right hand side of the equality you have quoted is a geometric series. Identifying the first term $a=1$ and the common ratio $r=1-\alpha$, the series converges to $a/(1-r) = 1 / \alpha$ if and only if $\vert r \vert < 1$. That is, it converges iff $0 < \alpha < 2$. Meaning the equality you’ve mentioned only holds if $\alpha$ falls within that range.
As I’ve written this without regard for the context of the time series technique you’ve mentioned, it might be worthwhile to check the implications of what this mathematical restriction on $\alpha$ means within that context.