# 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 $\tau(n)$?

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 $(l-x)^{-8}$."

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.