TITLE: Do finitely many plurigenera determine the Kodaira dimension?
QUESTION [21 upvotes]: Let $X$ be a smooth projective variety over a field of characteristic $0$ and let $K_X$ be the canonical bundle.
Recall that the Kodaira dimension $\kappa(X)$ is defined as the number $\kappa$ such that $$\alpha m^{\kappa}\le h^0(X,mK_X) \le \beta m^{\kappa}$$ for $\alpha,\beta>0$ and $m$ sufficiently large and divisible (or $\kappa=-\infty$ if $h^0(X,mK_X)=0$ for all $m$). It is well-known that $\kappa(X)$ is a birational invariant. A natural question is how large $m$ we need to take to determine $\kappa$. More precisely:

Is there an integer $M>0$, depending
  only $\dim X$, such that the values
  $h^0(X,K_X),h^0(X,2K_X),\ldots,h^0(X,MK_X)$
  determine $\kappa(X)$?

In particular, is there an $M>0$, depending on $\dim X$, such that $h^0(K_X)=\ldots=h^0(MK_X)=0$ implies that $\kappa=-\infty$?

REPLY [2 votes]: I guess in general the answer is NO. 
For instance if you take $X$ to be an $n$ dimensional variety which is $Y\times E$, where $E$ is an elliptic curve and $Y$ is an $n-1$ dimensional variety of general type with fast growing pluricanonical general, say $Y$ is a hypersurface of large degree $d$, then for any fixed $M$,  if you choose $d$ sufficiently large, $h^0(X,iK_X)$ can beat any sequence of numbers $a_i$ $(0\le i\le M)$ which you wrote for a fixed $n$-dimensional general type variety.