Limit with inverse trignometry [closed]

We have to find the value of $$\lim_{n\to \infty}\sum_{k=2}^n\arccos\left(\frac{1+\sqrt{(k-1)k(k+1)(k+2)}}{k(k+1)}\right)$$

I tried it a lot but could not find any start.

Please help me in this .

Solution 1:

HINT:

$$\cos^{-1}\left(\dfrac1k\cdot\dfrac1{k+1}+\sqrt{1-\dfrac1{k^2}}\cdot\sqrt{1-\dfrac1{(k+1)^2}}\right)=\cos^{-1}\dfrac1k-\cos^{-1}\dfrac1{k+1}$$

Use Telescoping series

Solution 2:

We have the identity $$\arccos \alpha -\arccos \beta = \arccos(\alpha \beta+\sqrt{1-\alpha^2}\sqrt{1-\beta^2})$$ Here we have that $\alpha =\frac{1}{k}$ and $\beta =\frac{1}{k+1}$. Then our limit transforms to $$\lim_{n\to \infty} \arccos(1/2)-\arccos(1/3) +\arccos(1/3)-\arccos(1/4)+\cdots$$ This limit can then be easily calculated. Hope it helps.