Inverse Tangent Difference Formula

Using

$$\sin^{-1}\left(\frac35\right)=\tan^{-1}\left(\frac34\right),$$

we get

$$\tan^{-1}\left(\frac34\right) -2\tan^{-1}\left(\frac12\right)$$

Now,

$$2\tan^{-1}\left(\frac12\right) = \tan^{-1}\left(\frac43\right)$$

Hence,

$$\tan^{-1}\left(\frac34\right) - \tan^{-1}\left(\frac43\right)$$

Using

$$\tan^{-1}a-\tan^{-1}b = \tan^{-1}\left(\frac{a-b}{1+ab}\right),$$

$$= \tan^{-1}\left( \frac{\frac34-\frac43}{1+\frac34\cdot\frac43} \right)$$

$$= \tan^{-1}\left( \frac{-\frac7{12}}{2} \right)$$

$$= \tan^{-1}\left(-\frac7{24}\right)$$

Therefore,

$$k=-\frac7{24}$$

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:

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

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

Let

$$T_r=\tan^{-1}\left(\frac{x}{x^2+r(r-1)}\right),\qquad r\ge2$$

Now,

$$\frac{x}{x^2+r(r-1)}=\frac{\frac{r}{x}-\frac{r-1}{x}}{1+\frac{r(r-1)}{x^2}}$$

Using

$$\tan^{-1}\left(\frac{A-B}{1+AB}\right)=\tan^{-1}A-\tan^{-1}B,$$

we get

$$T_r=\tan^{-1}\left(\frac{r}{x}\right)-\tan^{-1}\left(\frac{r-1}{x}\right)$$

Therefore,

$$S_n=\tan^{-1}\left(\frac1x\right)+\sum_{r=2}^{n}\left[\tan^{-1}\left(\frac{r}{x}\right)-\tan^{-1}\left(\frac{r-1}{x}\right)\right]$$

The series telescopes:

$$S_n=\tan^{-1}\left(\frac{n}{x}\right)$$

Taking

$$n\to\infty,$$

$$S_\infty=\lim_{n\to\infty}\tan^{-1}\left(\frac{n}{x}\right)$$

Since

$$x>0,$$

$$\frac{n}{x}\to\infty$$

Hence,

$$S_\infty=\frac{\pi}{2}$$

Comparing with $$\frac{\pi}{k},$$

$$k=2$$