Can the Y-combinator really terminate?
Terms containing the Y combinator can be weakly normalizing: they can terminate for a certain order of evaluation, i.e. for a certain evaluation strategy. No term containing the Y combinator applied to another term can be strongly normalizing, i.e. no term containing Y applied to another term terminates for all evaluation strategies, since the infinite reduction chain $Y \, M \to_\beta^* M \, (Y \, M) \to_\beta^* M \, (M \, (Y \, M)) \to_\beta^* \ldots$ is possible.
There are terms $M$ such that $Y \, M$ has no normal form, i.e. such that there is no reduction strategy that reduces $Y \, M$ to a value. For example $Y (\lambda x. x) \to_\beta^* (\lambda x.x) (Y (\lambda x.x)) \to_\beta Y (\lambda x. x)$ is a recursive functions that just spins (let x = x). (The lack of another reduction chain that terminates is left as an exercise for the reader.)
Under call-by-value (only ever reducing $M \, N$ when $M$ and $N$ are values), $Y \, M$ only converges if $M$ does not use its argument, i.e. if $Y \, M$ is not “really” recursive. Under call-by-name (always reducing the outermost redex first), $Y \, M$ converges if the recursive function that it defines terminates.