Let $\pi_n$ be the value of $\pi$ truncated to $n$ digits, e.g. $\pi_4=3.141$. I noticed that $\pi_n + \sin\pi_n$ approaches $\pi$ as $n$ grows. For example, $\sin\pi_4$ equals $0.0005926535551\dots$ which is $\pi - \pi_4$ accurate to 10 digits.
Is there a reason for this?
Note: the point of the question is not to ask why $\pi_n + \sin \pi_n \to \pi$ as $n \to \infty$, but rather, why the error of $|\pi_n + \sin \pi_n - \pi|$ is smaller than the error of $|\pi_n - \pi|$, and roughly quantify it.
To that end, here is a hint: define $\delta_n = \pi - \pi_n > 0$. Then $$\sin \pi_n = \sin (\pi - \delta_n) = \sin \delta_n.$$ Now consider the error $$\epsilon_n = -(\pi - \pi_n) + \sin \pi_n = -\delta_n + \sin \delta_n = -\delta_n + \delta_n - \frac{\delta_n^3}{3!} + O(\delta_n^5) = -\frac{\delta_n^3}{3!} + O(\delta_n^5).$$ So the difference of $\pi_n + \sin \pi_n$ from $\pi$ is on the order of $\delta_n^3 \sim 10^{-3n}$. That is to say, the number of correct digits of $\pi$ in the expression $\pi_n + \sin \pi_n$ is more than the number of correct digits of $\pi_n$, about three times as many.
Some relevant facts: $\sin\pi = 0$, and $\sin x$ is a continuous function of $x$. This tells us that $$\lim_{x\to\pi} \sin x = 0.$$
Clearly as $n\to\infty$, then $\pi_n \to \pi.$ After all, $|\pi_n - \pi| < 10^{-n}.$
So altogether you have
$$\lim_{n\to\infty} (\pi_n + \sin \pi_n) = \lim_{x\to\pi} (x + \sin x) = \left(\lim_{x\to\pi} x\right) + \left(\lim_{x\to\pi} \sin x\right) = \pi + 0 = \pi.$$
So you should expect $\pi_n + \sin \pi_n$ to converge to $\pi$ eventually. As you noticed, it seems to converge a lot faster than $\pi_n$ itself converges to $\pi$. That's because for $x \approx \pi$, $\sin x$ is an excellent approximation of $\pi - x$, and of course $x + (\pi - x) = \pi.$
A linear approximation of $y=\sin{x}$ near $\pi$ is $y=\pi - x$. This means that, near $\pi$, $\pi - x \approx \sin{x}$ and since your sequence approaches $\pi$ then we would expect the two values to approach each other.
If you graph the function $x+\sin x$, you'll find that the area around the point $(\pi,\pi)$ is flat. That is basically the reason behind this: if $x\approx\pi$, then $x+\sin x\approx\pi$ very closely (because the area around $\pi$ on the graph is almost flat).
