Status of $\tau(n)$ before Deligne Ramanujan's $\tau$ conjecture states that $$\tau(n)\sim O(n^{\frac{11}2+\epsilon})$$ which is a consequence of Deligne's proof of Weil conjectures. My question is what is best that could be proved possibly without etale cohomology regarding $\tau(n)$?
 A: In Hardy's
"Ramanujan -
Twelve Lectures on Subjects
Suggested by his
Life and Work",
in chapter 10,
section 10.7,
on the growth of
$\tau(n)$,
Hardy states that
"We saw in section 9.18 that
$\tau(n) = O(n^8)$.
Ramanujan gave a more
elementary proof that
$\tau(n) = O(n^7)$,
and this is the most
that has been proved by
"elementary" methods.
I
proved in 1918, by the method used by 
Littlewood and myself in our work
on Waring's problem, that
$\tau(n) = O(n^6)$.
Kloosterman proved in 1927 that
$\tau(n) = O(n^{47/8+c})$
for every positive $c$; 
Davenport and Salie proved independently in 1933 that
$\tau(n) = O(n^{35/6+c})$;
and finally Rankin proved in 1939 that
$\tau(n) = O(n^{29/5})$,
the best result yet known. 
The indices here are (apart from the $c$'s) 
less than $6$ by $1/8$, $1/6$, 
and
$1/5$, respectively."
Here is the proof
(from section 9.18,
page 156)
 that
$\tau(n) = O(n^8)$.
My apologies for any transcription errors.
"It follows from a formula of Jacobi 
which I have quoted several times
already that
$\sum \tau(n) x^n 
= x\{(1-x) (1-x^2)  \}^{24} 
= x(1-3x+ 5x^3 -7x^6+ ... )^8$,
the exponents in the series 
being the triangular numbers. 
Now $( 1- 3x + ... )^8$
is majorised by 
$\left(\sum_{n=0}^{\infty}
 (2n + 1) x^{n(n+1)/2}\right)^8$ ,
which is of order $( 1 - x )^{-8}$
 when $x\to 1$ (see below). 
Hence
$|\tau(n)|x^n
< \sum |\tau(n)| x^n
<A(1-x)^{-8}$,
where $A$ is a constant, 
for all $n$ and $x$. 
Taking $x = 1-1/n$, 
when $x^n$ is about
$1/e$, 
we find that
$\tau(n) = O(n^8)$."
Here is Hardy's proof
of the bound
$(1-x)^{-8}$:
"That of
$(\sum n x^{n^2/2})^8$
 or of
$(\int_0^{\infty} t e^{-y t^2/2} dt)^8$
, where $e^{-y} = x$. 
This is that of 
$y^{-8}$
 or $(l-x)^{-8}$."
A: The pentagonal number theorem implies that $\tau(n - 1)$ is at most the number of $24$-tuples of pentagonal numbers summing to $n$. There are $O(\sqrt{n})$ pentagonal numbers that can appear in this sum, which gives
$$\tau(n) \in O(n^{12}).$$
Edit: Consider also the following heuristic argument. The pentagonal number theorem in fact lets us write $\tau(n - 1)$ as a sum of $O(n^{12})$ signs. Assume that these signs are randomly distributed. Then one expects their sum to have absolute value $O(n^6)$ by a straightforward variance calculation. This is the same sort of argument that correctly suggests that Gauss sums should have absolute value around $\sqrt{p}$.
But in fact Ramanujan's conjecture is better than this by a factor of $\sqrt{n}$. I don't know where this extra savings comes from even heuristically.
