Let $N$ a large number and $P=P(N)$. We know that the "tail" of singular series of Goldbach problem is $$ \underset{q>P}{\sum}\,\frac{\mu(q)^{2}}{\phi(q)^{2}}\overset{q}{\underset{a=1}{\sum}^{*}}e\left(-N\frac{a}{q}\right).$$ (the symbol * indicates the condition $(a,q)=1$ ). What are the best estimates for it (conditional and unconditional)?
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Here is an unconditional error:
Restricting to summation over $q$ with $gcd(q,N)=n$ and using the formula for Ramanujan sums implies that the quantity you stated equals $$\sum_{n|N}\frac{\mu^2(n)}{\phi(n)} \sum_{m>P/n}\frac{\mu(m)}{\phi^2(m)},$$ where the summation is restricted to $m$ with $$\gcd(m,n)=\gcd(m,N/n)=1.$$ You should be able to show that $$\sum_{k>x}\frac{1}{\phi^2(k)}\ll \frac{1}{x}$$ and by bounding $\mu(n)$ trivially by $|\mu(n)|\leq 1$ this will give an error term for your problem which is $$O\Big(\frac{N^{\epsilon}}{P}\Big) (\forall \epsilon>0)$$ where the implied constant depends at most on $\epsilon.$. This method can also give more precise information instead of $N^\epsilon.$
About conditional errors:
The Riemann hypothesis is equivalent to $$\sum_{n \leq x}\mu(n)\ll_{\epsilon} x^{1/2+\epsilon} (\forall \epsilon>0).$$ One can then use this to show $$\sum_{k>x}\frac{\mu(k)}{\phi^2(k)}\ll_{\epsilon} \frac{1}{x^{3/2-\epsilon}}$$ and therefore the error term in your problem would be $$O(\frac{N^\epsilon}{P^{3/2-\epsilon}}),$$ i.e. you would be able to save half a power of $P$ under the Riemann hypothesis.
The difference in the two approaches is that in the first you bound the oscillating Mobius function trivially and in the second approach you take into account some cancellation in this oscillation. By the way the best unconditional estimate is achieved by using the bound $$\sum_{n \leq x}\mu(n)\ll x \exp(-c\sqrt(\log x))$$ where $c$ is a positive constant, which would then give an error term of the form $$O(N^{\epsilon}P^{-1} \exp(-c\sqrt(\log P))).$$
