Let $X$ be a smooth projective variety 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.

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

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 $\kappa=-\infty$?