On the parity of $\left\lfloor{\frac{3^n}{2^n}}\right\rfloor$

Let $a_n=(-1)^{\left\lfloor{\frac{3^n}{2^n}}\right\rfloor}$ and $$s_n=\sum_{k=1}^na_k.$$

Is it true that $s_n\le 0$ for all $n\geq 1$ ? (This is true for $n\le 100000$.)

In other words, odd numbers are always more than even numbers on the sequence $\left\lfloor{\frac{3^n}{2^n}}\right\rfloor$. This is unexpected, I think they should be roughly equal, and even numbers will exceed odd numbers sometimes.

Not a complete answer, but here are some results from computer simulation:

While the sequence $s_n$ starts out being negative for quite some time (as mentioned in the question, for more than the first $100\,000$ terms), it does reach positive values eventually. The first time this happens is for $n=331\,523$; there we have $165\,762$ even and $165\,761$ odd numbers. So the answer to the question would be no.

After that it continues to alternate between positive and negative values.

See this plot of $s_n$ for $n$ up to $1\,000\,000$:

Here is how it continues up to $5\,000\,000$:

Of course it would be nice if this could be verified without computer help... or to see a theoretical description of the asympotic behavior of $s_n$...