Question is
If the value of $\displaystyle \lim_{n\to\infty}\sum_{k=2}^{n}\cos^{-1}\left[\frac{1+\sqrt{(k-1)(k)(k+1)(k+2)}}{k(k+1)}\right]$ is equal to $\displaystyle\frac{120\pi}{m}$.
Then find the value of $m$.
What I did was to try putting the value of the variable $k$ form $1$ to infinity one by one and writing its correspondence value of $\arctan(x)$ And then tried to to connect all the terms. I am stuck at that point.. Any further help or any other better method would be of immense help.. Thanks in advance.....
Let $$\displaystyle S_{n} = \lim_{n\rightarrow \infty}\sum^{n}_{k=2}\cos^{-1}\left(\frac{1+\sqrt{(k-1)k(k+1)(k+2)}} {k(k+1)}\right)\;$$
Now Let $$\displaystyle T_{k} = \cos^{-1}\left[\frac{1}{k}\cdot \frac{1}{k+1}+\frac{\sqrt{(k-1)k(k+1)(k+2)}} {k(k+1)}\right]\;$$
Now Let $\displaystyle x=\frac{1}{k}$ and $\displaystyle y=\frac{1}{k+1}$
So $$\displaystyle \sqrt{1-x^2} = \sqrt{1-\frac{1}{k^2}} = \frac{\sqrt{(k-1)(k+1)}}{k}$$ and
$$\displaystyle \sqrt{1-y^2} = \sqrt{1-\frac{1}{(k+1)^2}} = \frac{\sqrt{k(k+2)}}{k+1}$$
So $\displaystyle T_{k}$ is of the form $$\cos^{-1}\left(xy+\sqrt{1-x^2}\sqrt{1-y^2}\right) = \cos^{-1}(y)-\cos^{-1}(x)\;,$$ because $(y<x)$
Now $$\displaystyle T_{k} = \cos^{-1}\left(\frac{1}{k+1}\right)-\cos^{-1}\left(\frac{1}{k}\right)$$
So $$\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=2}T_{k} = \lim_{n\rightarrow \infty}\left[\cos^{-1}\left(\frac{1}{n+1}\right)-\cos^{-1}\left(\frac{1}{2}\right)\right]$$
So we get $$\displaystyle \sum^{n}_{k=2}\cos^{-1}\left(\frac{1+\sqrt{(k-1)k(k+1)(k+2)}} {k(k+1)}\right)= \cos^{-1}(0)-\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2}-\frac{\pi}{3} = \frac{\pi}{6}$$
So $$\displaystyle S_{n} = \frac{\pi}{6} = \frac{120\pi}{m}\;,$$ We get $$m=720$$