Inverse Trigonometric Functions#0

Let $\displaystyle S_n = \sum_{k=1}^n \cot^{-1} \left(\frac {k(k+1)}4+4\right) = \cot^{-1} a_n$ . From the first few $a_n$ , we note that $a_n = \dfrac {n+17}{4n}$ . Let us prove in by induction that the claim is true for all $n \ge 1$ .

Proof: For $n=1$ , $a_1 = \dfrac {1+17}{4(1)} = \dfrac 92 = \dfrac 24 + 4$ as given. The claim is true for $n=1$ . Assuming the claim is true for $n$ , then:

$\begin{aligned} S_{n+1} & = S_n + \cot^{-1} \left(\frac {(n+1)(n+2)}4 + 4 \right) \\ & = \cot^{-1} a_n + \cot^{-1} \left(\frac {n^2+3n+18}4 \right) \\ & = \cot^{-1} \left(\frac {n+17}{4n} \right) + \cot^{-1} \left(\frac {n^2+3n+18}4 \right) \\ & = \cot^{-1} \left(\frac {\left(\frac {n+17}{4n} \right)\left(\frac {n^2+3n+18}4 \right)-1} {\left(\frac {n+17}{4n} \right)+\left(\frac {n^2+3n+18}4 \right)}\right) \\ & = \cot^{-1} \left(\frac {n^3+20n^2+65n+306}{4(n^3+3n^2+19n+17)}\right) \\ & = \cot^{-1} \left(\frac {(n+18)(n^2+2n+17)}{4(n+1)(n^2+2n+17)}\right) \\ & = \cot^{-1} \left(\frac {n+18}{4(n+1)}\right) \end{aligned}$

Therefore, the claim is also true for $n+1$ and hence true for all $n \ge 1$ .

Now, we have $\displaystyle \lim_{n \to \infty} S_n = \lim_{n \to \infty} \cot^{-1} \left(\frac {n+17}{4n}\right) = \lim_{n \to \infty} \cot^{-1} \left(\frac {1+\frac {17}n}4 \right) = \cot^{-1} \left(\dfrac 14\right) = \tan^{-1} 4$ .

Therefore, $p-q = 4-1 = \boxed{3}$ .

@Manu Sharma , could u pls explain how u solved this problem ?? Also(general doubt) how to solve such type of inverse+series problems ??