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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

${ \cos}^{ - 1} x - { \cos}^{ - 1} y = { \cos}^{ - 1}( xy + \sqrt{(1 - {x}^{2})(1 - {y}^{2} ) }$

TO FIND

$\displaystyle \lim_{n \to \infin} \sum\limits_{k=2}^{n} { \cos}^{ - 1} ( \frac{1 + \sqrt{(k - 1)k(k + 1)(k + 2)} }{k(k + 1)}$

EVALUATION

Here

$\displaystyle \sum\limits_{k=2}^{n} { \cos}^{ - 1} ( \frac{1 + \sqrt{(k - 1)k(k + 1)(k + 2)} }{k(k + 1)}$

$= \displaystyle \sum\limits_{k=2}^{n} { \cos}^{ - 1} ( \frac{1}{k(k + 1)} + \frac{ \sqrt{(k - 1)k(k + 1)(k + 2)} }{k(k + 1)}$

$= \displaystyle \sum\limits_{k=2}^{n} { \cos}^{ - 1} ( \frac{1}{k(k + 1)} + \sqrt{ \frac{ (k - 1)k(k + 1)(k + 2) }{ {k}^{2} {(k + 1)}^{2} } }$

$= \displaystyle \sum\limits_{k=2}^{n} { \cos}^{ - 1} ( \frac{1}{k(k + 1)} + \sqrt{ \frac{ ( {k}^{2} - 1)( {k}^{2} + 2k) }{ {k}^{2} {(k + 1)}^{2} } }$

$= \displaystyle \sum\limits_{k=2}^{n} { \cos}^{ - 1} ( \frac{1}{k(k + 1)} + \sqrt{ \frac{ ( {k}^{2} - 1) \{ {(k + 1)}^{2} - 1 \} }{ {k}^{2} {(k + 1)}^{2} } }$

$= \displaystyle \sum\limits_{k=2}^{n} { \cos}^{ - 1} ( \frac{1}{k(k + 1)} + \sqrt{ ( 1- \frac{1}{ {k}^{2} } )(1 - \frac{1}{ {(k + 1)}^{2} }) }$

$= \displaystyle \sum\limits_{k=2}^{n} ( { \cos}^{ - 1} \frac{1}{k+1} - { \cos}^{ - 1} \frac{1}{k } )$

$= \displaystyle ({ \cos}^{ - 1} \frac{1}{3} - { \cos}^{ - 1} \frac{1}{2} + { \cos}^{ - 1} \frac{1}{4} - { \cos}^{ - 1} \frac{1}{3} + ..... + { \cos}^{ - 1} \frac{1}{n+1} - { \cos}^{ - 1} \frac{1}{n })$

$= \displaystyle ({ \cos}^{ - 1} \frac{1}{n+1} - { \cos}^{ - 1} \frac{1}{2})$

So

$\displaystyle \lim_{n \to \infin} \sum\limits_{k=2}^{n} { \cos}^{ - 1} ( \frac{1 + \sqrt{(k - 1)k(k + 1)(k + 2)} }{k(k + 1)}$

$= \displaystyle \lim_{n \to \infin} ({ \cos}^{ - 1} \frac{1}{n+1} - { \cos}^{ - 1} \frac{1}{2})$

$= \displaystyle ( { \cos}^{ - 1} 0 - { \cos}^{ - 1} \frac{1}{2} )$

$= \displaystyle \frac{\pi}{2} - \frac{\pi}{3}$

$= \displaystyle \: \frac{\pi}{6}$

Answer 2

Given,

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\sum_{k=2}^n\cos^{-1}\left(\dfrac{1+\sqrt{(k-1)k(k+1)(k+2)}}{k(k+1)}\right)\quad\quad\dots(1)$

We know the formula,

$\longrightarrow \cos^{-1}x-\cos^{-1}y=\cos^{-1}\left(xy+\sqrt{(1-x^2)(1-y^2)}\right)$

Replacing $x=\dfrac{1}{a}$ and $y=\dfrac{1}{b},$

$\longrightarrow \cos^{-1}\left(\dfrac{1}{a}\right)-\cos^{-1}\left(\dfrac{1}{b}\right)=\cos^{-1}\left(\dfrac{1}{a}\cdot\dfrac{1}{b}+\sqrt{\left(1-\dfrac{1}{a^2}\right)\left(1-\dfrac{1}{b^2}\right)}\right)$

$\longrightarrow \cos^{-1}\left(\dfrac{1}{a}\right)-\cos^{-1}\left(\dfrac{1}{b}\right)=\cos^{-1}\left(\dfrac{1}{ab}+\sqrt{\left(\dfrac{a^2-1}{a^2}\right)\left(\dfrac{b^2-1}{b^2}\right)}\right)$

$\longrightarrow \cos^{-1}\left(\dfrac{1}{a}\right)-\cos^{-1}\left(\dfrac{1}{b}\right)=\cos^{-1}\left(\dfrac{1}{ab}+\sqrt{\dfrac{(a^2-1)(b^2-1)}{a^2b^2}}\right)$

$\longrightarrow \cos^{-1}\left(\dfrac{1}{a}\right)-\cos^{-1}\left(\dfrac{1}{b}\right)=\cos^{-1}\left(\dfrac{1+\sqrt{(a^2-1)(b^2-1)}}{ab}\right)\quad\quad\dots(2)$

We see that,

$\longrightarrow\dfrac{1+\sqrt{(k-1)k(k+1)(k+2)}}{k(k+1)}=\dfrac{1+\sqrt{(k-1)(k+1)k(k+2)}}{k(k+1)}$

$\longrightarrow\dfrac{1+\sqrt{(k-1)k(k+1)(k+2)}}{k(k+1)}=\dfrac{1+\sqrt{(k^2-1)(k^2+2k)}}{k(k+1)}$

$\longrightarrow\dfrac{1+\sqrt{(k-1)k(k+1)(k+2)}}{k(k+1)}=\dfrac{1+\sqrt{(k^2-1)(k^2+2k+1-1)}}{k(k+1)}$

$\longrightarrow\dfrac{1+\sqrt{(k-1)k(k+1)(k+2)}}{k(k+1)}=\dfrac{1}{k(k+1)}+\dfrac{\sqrt{(k^2-1)((k+1)^2-1)}}{k(k+1)}$

Comparing this with (2) we get,

• $a=k$

• $a=k+1$

And, for some $a<b,$

$\longrightarrow \cos^{-1}a>\cos^{-1}b$

Thus,

$\longrightarrow k+1>k$

$\longrightarrow\dfrac{1}{k+1}<\dfrac{1}{k}$

$\longrightarrow\cos^{-1}\left(\dfrac{1}{k+1}\right)>\cos^{-1}\left(\dfrac{1}{k}\right)$

Therefore,

$\longrightarrow\cos^{-1}\left(\dfrac{1}{k+1}\right)-\cos^{-1}\left(\dfrac{1}{k}\right)=\cos^{-1}\left(\dfrac{1}{k(k+1)}+\dfrac{\sqrt{(k^2-1)((k+1)^2-1)}}{k(k+1)}\right)$

Then (1) becomes,

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\sum_{k=2}^n\left(\cos^{-1}\left(\dfrac{1}{k+1}\right)-\cos^{-1}\left(\dfrac{1}{k}\right)\right)$

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\sum_{k=2}^n\left(\cos^{-1}\left(\dfrac{1}{k+1}\right)\right)-\lim_{n\to\infty}\sum_{k=2}^n\left(\cos^{-1}\left(\dfrac{1}{k}\right)\right)\quad\quad\dots(3)$

Now each summation is written in this way.

$\displaystyle\longrightarrow\sum_{k=2}^n\left(\cos^{-1}\left(\dfrac{1}{k+1}\right)\right)=\sum_{k=2}^{n-1}\left(\cos^{-1}\left(\dfrac{1}{k+1}\right)\right)+\left(\cos^{-1}\left(\dfrac{1}{n+1}\right)\right)$

$\displaystyle\longrightarrow\sum_{k=2}^n\left(\cos^{-1}\left(\dfrac{1}{k+1}\right)\right)=\sum_{k=3}^n\left(\cos^{-1}\left(\dfrac{1}{k}\right)\right)+\left(\cos^{-1}\left(\dfrac{1}{n+1}\right)\right)$

Here last term is taken out of the summation.

$\displaystyle\longrightarrow\sum_{k=2}^n\left(\cos^{-1}\left(\dfrac{1}{k}\right)\right)=\left(\cos^{-1}\left(\dfrac{1}{2}\right)\right)+\sum_{k=3}^n\left(\cos^{-1}\left(\dfrac{1}{k}\right)\right)$

Here first term is taken out of the summation.

Then (3) becomes,

$\begin{aligned}\displaystyle\longrightarrow L=\ \ &\lim_{n\to\infty}\left[\sum_{k=3}^n\left(\cos^{-1}\left(\dfrac{1}{k}\right)\right)+\left(\cos^{-1}\left(\dfrac{1}{n+1}\right)\right)\right]\\-\ \ &\lim_{n\to\infty}\left[\left(\cos^{-1}\left(\dfrac{1}{2}\right)\right)+\sum_{k=3}^n\left(\cos^{-1}\left(\dfrac{1}{k}\right)\right)\right]\end{aligned}$

$\begin{aligned}\displaystyle\longrightarrow L=\ \ &\lim_{n\to\infty}\sum_{k=3}^n\left(\cos^{-1}\left(\dfrac{1}{k}\right)\right)+\lim_{n\to\infty}\left(\cos^{-1}\left(\dfrac{1}{n+1}\right)\right)\\-\ \ &\lim_{n\to\infty}\left(\cos^{-1}\left(\dfrac{1}{2}\right)\right)-\lim_{n\to\infty}\sum_{k=3}^n\left(\cos^{-1}\left(\dfrac{1}{k}\right)\right)\end{aligned}$

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left(\cos^{-1}\left(\dfrac{1}{n+1}\right)\right)-\lim_{n\to\infty}\left(\cos^{-1}\left(\dfrac{1}{2}\right)\right)$

$\displaystyle\longrightarrow L=\left(\cos^{-1}\left(\dfrac{1}{\infty}\right)\right)-\left(\cos^{-1}\left(\dfrac{1}{2}\right)\right)$

$\displaystyle\longrightarrow L=\cos^{-1}0-\cos^{-1}\left(\dfrac{1}{2}\right)$

$\displaystyle\longrightarrow L=\dfrac{\pi}{2}-\dfrac{\pi}{3}$

$\displaystyle\longrightarrow\underline{\underline{L=\dfrac{\pi}{6}}}$