Let $$S_k = \sum_{r=1}^{k} \tan^{-1}\left(\frac{6^r}{2^{2r+1} + 3^{2r+1}}\right)$$, then $$\lim_{k \to \infty} S_k$$ is equal to:

We need to evaluate $$S_k = \sum_{r=1}^{k} \tan^{-1}\left(\frac{6^r}{2^{2r+1} + 3^{2r+1}}\right)$$ and find $$\lim_{k \to \infty} S_k$$.

We write $$6^r = 2^r \cdot 3^r$$ and note that $$2^{2r+1} + 3^{2r+1} = 2 \cdot 4^r + 3 \cdot 9^r$$. We claim each term telescopes as $$\tan^{-1}\left(\frac{3}{2}\right)^{r+1} - \tan^{-1}\left(\frac{3}{2}\right)^r$$.

To verify, we use the subtraction formula for inverse tangent: $$\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A - B}{1 + AB}\right)$$ (when $$AB > -1$$). Setting $$A = (3/2)^{r+1}$$ and $$B = (3/2)^r$$, we get the numerator $$A - B = (3/2)^r\left(\frac{3}{2} - 1\right) = \frac{1}{2}(3/2)^r$$ and the denominator $$1 + AB = 1 + (3/2)^{2r+1}$$.

So the difference equals $$\tan^{-1}\left(\frac{(3/2)^r / 2}{1 + (3/2)^{2r+1}}\right)$$. To match with our original expression, we multiply both the numerator and denominator inside by $$2^{2r+1}$$. The numerator becomes $$(3/2)^r \cdot 2^{2r} = 3^r \cdot 2^r = 6^r$$, and the denominator becomes $$2^{2r+1} + (3/2)^{2r+1} \cdot 2^{2r+1} = 2^{2r+1} + 3^{2r+1}$$. This confirms the telescoping identity.

Therefore, $$S_k = \sum_{r=1}^{k}\left[\tan^{-1}\left(\frac{3}{2}\right)^{r+1} - \tan^{-1}\left(\frac{3}{2}\right)^r\right] = \tan^{-1}\left(\frac{3}{2}\right)^{k+1} - \tan^{-1}\left(\frac{3}{2}\right)$$.

As $$k \to \infty$$, $$(3/2)^{k+1} \to \infty$$, so $$\tan^{-1}\left((3/2)^{k+1}\right) \to \frac{\pi}{2}$$. Therefore, $$\lim_{k \to \infty} S_k = \frac{\pi}{2} - \tan^{-1}\left(\frac{3}{2}\right)$$. Using the identity $$\frac{\pi}{2} - \tan^{-1}(\theta) = \cot^{-1}(\theta)$$, we get $$\lim_{k \to \infty} S_k = \cot^{-1}\left(\frac{3}{2}\right)$$.