Step 1: Understanding the series. The given series is: \[ \cot^{-1} \left( \frac{7}{4} \right) + \cot^{-1} \left( \frac{19}{4} \right) + \cot^{-1} \left( \frac{39}{4} \right) + \cot^{-1} \left( \frac{67}{4} \right) + \cdots \] This is a standard arccotangent series.
Step 2: Identifying the pattern. The series consists of terms of the form: \[ \cot^{-1} \left( \frac{4n + 3}{4} \right), \quad n = 1, 2, 3, \dots \] By using the identity for the sum of arccotangents: \[ \cot^{-1}(x) + \cot^{-1}(y) = \cot^{-1} \left( \frac{xy - 1}{x + y} \right) \] we can simplify the series.
Step 3: Applying the formula to the series. By applying the identity iteratively and simplifying, we can find that the sum of the infinite series converges to: \[ \pi - \cot^{-1} \left( \frac{1}{2} \right) \]
Step 4: Conclusion. The sum of the infinite series is: \[ \boxed{\pi - \tan^{-1} \left( \frac{1}{2} \right)} \] Final Answer: \[ \boxed{4}. \]
If the sum of the first 10 terms of the series \[ \frac{4 \cdot 1}{1 + 4 \cdot 1^4} + \frac{4 \cdot 2}{1 + 4 \cdot 2^4} + \frac{4 \cdot 3}{1 + 4 \cdot 3^4} + \ldots \] is \(\frac{m}{n}\), where \(\gcd(m, n) = 1\), then \(m + n\) is equal to _____.
If \(\sum\)\(_{r=1}^n T_r\) = \(\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}\) , then \( \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} \) is equal to :
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: