What upper bounds on $\sum_{nWhat are the current best known upper bounds on
$$\sum_{n<x}a_n$$
where $a_n$ are defined implicitly by $L(E,s)=\sum_{n=1}^{\infty}\frac{a_n}{n^s}$ where $L(E,s)$ is an Elliptic L Function and $E/\mathbb{Q}$ is an elliptic curve over $\mathbb{Q}$? The notion of "best" might be a bit fuzzy here since on one hand we want a bound that grows slowly with respect to $x$, but on the other we can a bound which goes not get too large as invariants of the curve (such as the conductor) get large.
I would expect something of the type $\left|\sum_{n<x}a_n\right|\ll_E x\log(x)^r$ where $r$ is the rank, judging by the BSD conjecture being initially stated in the form $\prod_{p<x}\frac{N_p}{p}\sim C\log(x)^r$, but the issue is that the BSD conjecture remains a conjecture and there is no immediately obvious way to find an upper bound on $C$ in terms of invariants of the curve.
Any bounds of the quantity $\sum_{n<x}a_n$ or adjacent objects would be highly appreciated. Thank you.
 A: The Weil bound $a_n \leq  d_2(n)\sqrt{n}$ where $n$ is the divisor function gives $ \sum_{n < x} a_n < \sum_{n_1n_2<x} \sqrt{n_1n_2} = O ( x^{3/2} \log x )$. This is the trivial bound in this setting.
To do better than that, the main approach will be to use the modularity theorem, which recognizes $a_n$ as Fourier coefficients of a modular form of weight $2$ and level $N$ equal to the conductor of the elliptic curve.
For $x$ large compared to the conductor $N$, we can do better directly using this modularity, applying the Voronoi summation formula to obtain a bound whose exact form I don't remember off the top of my head but which will have shape $O( x^{1/2 + \epsilon} N^{1/2})$.
You can combine these to get a bound like $O ( x^{1 + \epsilon} N^{1/4})$, which is technically of the form you desire, but whose dependence on $N$ is probably worse than you want.
Doing better than this will be closely related to bounds on the $L$-function of the elliptic curve. In particular, you can do a little better when $x$ is close to $N^{1/2}$ using subconvexity results.
Getting the dependence on $x$ all the way to $  x^{1+\epsilon}$ while keeping the dependence on $N$ small, say $N^{\epsilon}$, is problematic, because it implies
$$L( E, s) = \sum_{n=1}^{\infty} a_n n^{-s} = \sum_{d =2}^{\infty} \left(\sum_{n < d} a_n\right) \left(  (d-1)^{-s} - d^{-s} \right)$$ $$ \ll N^\epsilon \sum_{d=2}^{\infty} d \log (d)^r \left(  (d-1)^{-s} - d^{-s} \right) ,$$ which, because that sum converges for $s>1$, gives an $O(n^\epsilon)$ estimate of $L(E,s)$ for any $s>1$, which is equivalent to the Lindelöf hypothesis for this $l$-function.
I don't think you will see the rank of the curve playing a big role in this sum. The rank shows up in a sum over primes, which is related to the logarithmic derivative of the $L$-function, much more than it does in this sum, which is related to the $L$-function itself, because zeroes become poles when you take a logarithmic derivative.
