Inverse Trigonometric Functions#1

arccot[(r^2+3r+2)+1]=arccot[(r+2)(r+1)+1] which is terminating series now as shown below therefore Sn= arccot(r+1)-arccot(r+2).rest is simple

From the first few $n$ , we note that $S_n = \cot^{-1} \left(\dfrac {2n+5}n\right)$ . Let us prove by induction that the claim is true for all $n\ge 1$ .

Proof: For $n=1$ , $S_1 = \cot^{-1}\left(1^2+3(1)+3\right) = \cot^{-1} \left(\dfrac {2(1)+5}1\right) = 7$ . Therefore, the claim is true for $n=1$ . Assuming the claim is true for $n$ , then:

$\begin{aligned} S_{n+1} & = S_n + \cot^{-1} \left((n+1)^2 + 3(n+1)+3\right) \\ & = \cot^{-1} \left(\frac {2n+5}n \right) + \cot^{-1} \left(n^2 + 5n+7\right) \\ & = \cot^{-1} \left(\frac {\left(\frac {2n+5}n \right)\left(n^2 + 5n+7\right)-1}{\frac {2n+5}n + n^2 + 5n+7} \right) \\ & = \cot^{-1} \left(\frac {2n^3+15n^2+38n+35}{n^3+5n^2+9n+5} \right) \\ & = \cot^{-1} \left(\frac {(2n+7)(n^2+4n+5)}{(n+1)(n^2+4n+5)} \right) \\ & = \cot^{-1} \left(\frac {2(n+1)+5}{n+1} \right) \end{aligned}$

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

And we have:

$\begin{aligned} S_5 & = \frac {2(5)+5}5 = \color{#3D99F6} 3 & \color{#3D99F6} \text{True} \\ S_6 & = \frac {2(6)+5}6 = \color{#3D99F6} \frac {17}6 & \color{#3D99F6} \text{True} \\ S_8 & = \frac {2(8)+5}8 = \color{#D61F06} \frac {21}8 \ne 5 & \color{#D61F06} \text{False} \\ S_\infty & = \lim_{n\to \infty} \frac {2n+5}n = \lim_{n\to \infty} 2 + \frac 5n = \color{#3D99F6} 2 & \color{#3D99F6} \text{True} \end{aligned}$