$$\lim_{n \to \infty} \tan \sum_{r=1}^{n} \tan^{-1}\frac{1}{1 + r + r^2}$$ is equal to ______.

We need to evaluate $$\lim_{n \to \infty} \tan\left(\sum_{r=1}^{n} \tan^{-1}\frac{1}{1 + r + r^2}\right)$$.

We observe that $$\frac{1}{1 + r + r^2} = \frac{1}{1 + r(r+1)} = \frac{(r+1) - r}{1 + r(r+1)}$$.

Using the identity $$\tan^{-1} A - \tan^{-1} B = \tan^{-1}\frac{A - B}{1 + AB}$$, we can write $$\tan^{-1}\frac{(r+1) - r}{1 + r(r+1)} = \tan^{-1}(r+1) - \tan^{-1}(r)$$.

So the sum telescopes: $$\sum_{r=1}^{n} \left[\tan^{-1}(r+1) - \tan^{-1}(r)\right] = \tan^{-1}(n+1) - \tan^{-1}(1)$$.

As $$n \to \infty$$, $$\tan^{-1}(n+1) \to \frac{\pi}{2}$$ and $$\tan^{-1}(1) = \frac{\pi}{4}$$.

So the sum approaches $$\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$.

Therefore, $$\lim_{n \to \infty} \tan\left(\sum_{r=1}^{n} \tan^{-1}\frac{1}{1 + r + r^2}\right) = \tan\frac{\pi}{4} = 1$$.

Hence, the answer is $$1$$.