$\newcommand{\spec}{\mathrm{Spec}} \newcommand{\ra}{\rightarrow} \newcommand{\oh}{\mathcal{O}} \newcommand{\P}{\mathbb{P}}$
Let $f : X \ra \spec(A)$ be a projective morphism (Hartshorne's definition) with $A$ a noetherian ring and let $F \in \mathsf{Coh}(X)$. Let $L$ be an ample line bundle. Then there is $n_0 \geq 0$ such that $R^kf_*(F \otimes L^{\otimes n})=0$ for all $n \geq n_0$ and $k>0$.
In my class notes we have proved this by showing that $F(n) = F \otimes \oh(n)$ has no cohomology in positive degrees for $n$ sufficiently large, i.e. that $H^k(X, F(n))=0$ for $k>0$ and $n$ sufficiently large (this is where the proof in the class notes stops). This is what Hartshorne proves in p. 228 of his book, but the original claim (Theorem 5.2, p. 228) is that $H^k(X, F(n))=0$ rather than the statement about higher pushforwards $$R^kf_*(F \otimes L^{\otimes n})=0.$$ I'd like to know how $R^kf_*(F \otimes L^{\otimes n})=0$ follows from the vanishing cohomology result.
I've tried writing $R^kf_*(F \otimes L^{\otimes n}) = H^k(f_*(I^\bullet))$ where $I^\bullet$ is an injective resolution of $F \otimes L^{\otimes n}$, but I don't see how to relate the higher pushforward to the higher global sections functor which is what sheaf cohomology is)?
Thank you.
Edit: I've just realised that my question is perhaps related to a question I asked a few days ago which says that under certain assumptions $R^pf_*\mathcal{F} \cong \widetilde{H^p(X, \mathcal{F})}$; one of those assumptions is that $X$ is noetherian. I think this is OK - am I correct in saying that $\P^r_A$ is noetherian? It has a standard cover by affines $\mathbb{A}^m$, and each of those are noetherian because their associated rings are polynomial rings which are noetherian.
With this all being said, it seems that this is why/how we can make a connection between pushforwards and global sections/sheaf cohomology (I think maybe we should also note that the tilde functor $\widetilde{-}$ is exact), though the proof of $R^pf_*\mathcal{F} \cong \widetilde{H^p(X, \mathcal{F})}$ in my other question is rather technical, so I'm wondering if there is a simpler way to answer my original question?
To summarise my edit, is the way I've outlined above using my other question from a few days ago a valid way of answering my original question in this post? If it is valid, is there a quicker/more obvious way of answering my original question?
Thanks again.
One sentence answer would be look up Grothendieck complex and semi-continuity. Under your hypothesis there is a finite complex of $A$-projective modules, $P_.: 0\to P_0\to P_1\to\cdots\to P_n\to 0$ such that for any $Y\to \mathrm{Spec}\, A$, if $g:X\times_A Y\to Y$ is the induced morphism and $G$ is the pull back of $F$ to $X\times_A Y$, $R^kg_*G$ is the $k^{th}$ cohomolgy (as sheaves) of the pull back of $P_.$ to $Y$.
Now, apply this to $F\otimes L^n$ where $n\gg 0$, so that $H^k(X\otimes k(x), F\otimes L^n)=0$ for all $x\in X$ and all $k>0$. This says the corresponding $P_.$ is exact and thus $R^kf_*(F\otimes L^n)=0$ for all $k>0$.