Let $j, k, n\in \mathbb{N}$ and $\tau(k)$ be the Ramanujan tau function and let $$S(n,j)=\frac{1}{\tau(n) n^j}\sum_{k=1}^n \tau(k) k^j. \tag{1}$$ Numerical results for several values of $j, n$ showed that $S(n,j)>0$. Thus we are led to believe that $$S(n,j)>0 \tag{2}$$ To remove the potential rare possibility of $\tau(n)=0$ for some $n\in\mathbb{N}$, we would like to replace $n$ with $7^n$ and try to prove $$S(7^n,j)>0 \tag{3}.$$
Does there exist a counter example or a proof?
This problem showed up when we study the zeros of the L function associated with the Ramanujan tau-function: $$L_{\tau}(s)=\sum_{k=1}^\infty \frac{\tau(k)}{ k^s}\quad Re(s)>\frac{13}{2}\tag{4}$$
Thanks
-mike