Theorem on Formal Function Hartshorne Let consider the Thm on Formal Functions [see Hartshorne, III.11.1]. We have with $f:X \to Y$ a projective morphism of noetherian schemes,  $\mathcal{F}$ be a coherent sheaf on $X$ and let $y\in Y$. Then the natural map 
$$
R^i f_\ast (\mathcal{F})_y^\hat{} \to \varprojlim H^i(X_n, \mathcal{F}_n)
$$
is an isomorphism for all $i \geq 0$. 
My question is from what $n$ does the inverse limit start? From $n=1$ or $n=0$?
The background of my question is that all $\mathcal{F}_n $ come from the exact sequence $$\displaystyle  0 \rightarrow \mathcal{I}^n  \mathcal{F} \rightarrow \mathcal{F} \rightarrow \mathcal{F}_n  \rightarrow 0, $$ 
And by $\mathcal{I}^0\mathcal{F}$ it would imply that $\mathcal{F}_0 =\mathcal{F} $ and therefore the limit would map surjectively on global sections $H^i(X_n, \mathcal{F})$ what would be absurd for example if we consider the structure sheaf $\mathcal{F}= \mathcal{O}_X$.
 A: It doesn't matter; you can restrict to any infinite subset of natural numbers and get the same limit. (see final functor for the general notion)
In general, the canonical 'projections'
$$ \left(\varprojlim F_\bullet \right) \to F_n $$ 
need not be epimorphisms. So, if you find yourself in a situation where it would be absurd for one of these maps to be an epimorphism... that simply isn't a problem at all.
As an example, consider the family of subsets of $\mathbb{R}$ defined by $S_n = (-\infty, \frac{1}{n})$, where the transition maps are all inclusions. The inverse limit of these sets is simply their intersection:
$$ \left(\varprojlim S_\bullet\right) = (-\infty, 0] $$ 
and the canonical projections are simply inclusion. And clearly, none of the inclusions $(-\infty, 0] \to (-\infty, \frac{1}{n})$ are surjective.
A: The answer is in Hartshorne, right at the bottom of p. 276: $n \geq 1$. 
(You want $\mathrm{Spec}\,\mathcal O_y/\mathfrak m^n_y$ to be topologically just a point, and this happens starting from $n = 1$.)
