# 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 $\tau(n)$?

#### Solutions Collecting From Web of "Status of $\tau(n)$ before Deligne"

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
$\sum \tau(n) x^n = x\{(1-x) (1-x^2) \}^{24} = x(1-3x+ 5x^3 -7x^6+ … )^8$,
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 $(l-x)^{-8}$.”

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.