Direct Composition of $$\tan$$ and $$\tan^{-1}$$

The general term is $$T_n = \tan^{-1}\left(\frac{1}{n^2 + 3n + 3}\right)$$.

We can manipulate the denominator to fit the standard telescoping form $$1 + xy$$:

$$n^2 + 3n + 3 = 1 + (n^2 + 3n + 2) = 1 + (n+1)(n+2)$$

Notice that the difference between the factors $$(n+2)$$ and $$(n+1)$$ is exactly $$1$$:

$$T_n = \tan^{-1}\left(\frac{(n+2) - (n+1)}{1 + (n+1)(n+2)}\right)$$

Using the identity $$\tan^{-1}\left(\frac{x-y}{1+xy}\right) = \tan^{-1}(x) - \tan^{-1}(y)$$:

$$T_n = \tan^{-1}(n+2) - \tan^{-1}(n+1)$$

$$S = \left[\tan^{-1}(3) - \tan^{-1}(2)\right] + \left[\tan^{-1}(4) - \tan^{-1}(3)\right] + \dots + \left[\tan^{-1}(9) - \tan^{-1}(8)\right]$$

All intermediate terms cancel out, leaving:

$$S = \tan^{-1}(9) - \tan^{-1}(2)$$

$$\tan(S) = \tan\left(\tan^{-1}(9) - \tan^{-1}(2)\right) = \frac{9 - 2}{1 + (9)(2)} = \frac{7}{19}$$

$$19\tan(S) = 19 \times \frac{7}{19} = 7$$