Consider the offset logarithmic integral which approximates the number of primes up to a given $x$ quite well $$ \mathrm{Li}(x)=\int_2^x\frac{1}{\log t}~dt$$
And consider the alternating series
$$\mathrm{Z}(x)= \sum_{n=2}^x (-1)^{n+1}e^{\sin\big(\frac{(-1)^{n+1}}{\log n}\big)} $$
$\mathrm{Z}(x)$ remarkably stays close to $\mathrm{Li}(x)$ and is at least asymptotic to $\mathrm{Li}(x)$ from what I can tell. For example I tested this at $x=10^8$ the difference between the two is only about $4$ with $\mathrm{Z}(x)$ slightly less than the offset logarithmic integral.
What is the order of $\mathrm{Z}(x)-\pi(x)?$
Notice that $\sin z=z+O(z)$ and $e^z=1+z+\frac12z^2+O(z^3)$ as $z\to0$, so we have
\begin{aligned} Z(x) &=\sum_{2\le n\le x}(-1)^{n+1}e^{(-1)^{n+1}/\log n+O(1/\log^3n)} \\ &=\sum_{2\le n\le x}(-1)^{n+1}\left[1+{(-1)^{n+1}\over\log n}+{1\over2\log^2n}+O\left(1\over\log^3n\right)\right] \\ &=\sum_{2\le n\le x}{1\over\log n}+\sum_{2\le n\le x}(-1)^{n+1}+\frac12\sum_{2\le n\le x}{(-1)^{n+1}\over\log^2n}+O\left(\sum_{2\le n\le x}{1\over\log^3n}\right). \end{aligned}
Notice that the second sum is bounded, the third sum converges, and
$$ \sum_{2\le n\le x}{1\over\log^3n}\le\sum_{2\le n\le\sqrt x}{1\over\log^32}+\sum_{\sqrt x<n\le x}{1\over\log^3\sqrt x}=O\left(x\over\log^3x\right), $$
so we have
$$ Z(x)=\sum_{2\le n\le x}{1\over\log n}+O\left(x\over\log^3x\right).\tag1 $$
It can be proven that the remaining summation has bounded difference from $\operatorname{Li}(x)$, so we conclude that
$$ Z(x)-\operatorname{Li}(x)=O\left(x\over\log^3x\right).\tag2 $$
Although (2) only provides an O-bound, if we expand more terms from $\sin z$ and $e^z$, it is pretty likely that one will conclude the true order of error is $x/(\log x)^A$ for some fixed $A\ge3$, which is much worse than the order of $\pi(x)-\operatorname{Li}(x)$:
$$ \pi(x)-\operatorname{Li}(x)=O(xe^{-(\log x)^B}) $$
for some $0<B<1$.