Inverse Tangent Difference Identity

The general term of the series is:

$$T_n = \tan^{-1}\left(\frac{2n}{n^4 + n^2 + 2}\right)$$

We can rewrite the denominator to expose the standard form $$1 + ab$$:

$$n^4 + n^2 + 2 = 1 + (n^4 + n^2 + 1)$$

Using the algebraic factorization $$(n^4 + n^2 + 1) = (n^2 + n + 1)(n^2 - n + 1)$$, the term becomes:

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

We can rewrite the numerator $$2n$$ as the difference of these two factors:

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

Substituting this into the expression for $$T_n$$:

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

Using the identity $$\tan^{-1}\left(\frac{a - b}{1 + ab}\right) = \tan^{-1}a - \tan^{-1}b$$, we get:

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

Let $$V_n = \tan^{-1}(n^2 + n + 1)$$. Notice that:

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

Thus, the general term telescopes as:

$$T_n = V_n - V_{n-1}$$

The partial sum $$S_k$$ up to $$k$$ terms is:

$$S_k = \sum_{n=1}^{k} (V_n - V_{n-1}) = V_k - V_0$$

Substituting the definitions back in:

$$S_k = \tan^{-1}(k^2 + k + 1) - \tan^{-1}(0^2 + 0 + 1)$$

$$S_k = \tan^{-1}(k^2 + k + 1) - \tan^{-1}(1)$$

Taking the limit as $$k \to \infty$$ for the infinite sum:

$$S = \lim_{k \to \infty} \left[\tan^{-1}(k^2 + k + 1) - \tan^{-1}(1)\right]$$

$$S = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$