Question

$\displaystyle \rm \lim_{n \to \infty } \sum_{r = 1}^n {cot}^{ - 1} \left ( \frac{ {r}^{3} - r + \dfrac{1}{r} }{2} \right)$​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\rm{\lim_{n\to\infty}\sum^{n}_{r=1}\,cot^{-1}\left(\dfrac{{r}^{3}-r+\dfrac{1}{r}}{2}\right)}$

$\displaystyle\rm{=\lim_{n\to\infty}\sum^{n}_{r=1}\,tan^{-1}\left(\dfrac{2}{{r}^{3}-r+\dfrac{1}{r}}\right)}$

$\displaystyle\rm{=\lim_{n\to\infty}\sum^{n}_{r=1}\,tan^{-1}\left(\dfrac{2r}{{r}^{4}-{r}^{2}+1}\right)}$

$\displaystyle\rm{=\lim_{n\to\infty}\sum^{n}_{r=1}\,tan^{-1}\left(\dfrac{2r}{1+{r}^{2}\left({r}^{2}-1\right)}\right)}$

$\displaystyle\rm{=\lim_{n\to\infty}\sum^{n}_{r=1}\,tan^{-1}\left\{\dfrac{2r}{1+r(r+1)\cdot\,r(r-1)}\right\}}$

$\displaystyle\rm{=\lim_{n\to\infty}\sum^{n}_{r=1}\,tan^{-1}\left\{\dfrac{r(r+1)-r(r-1)}{1+r(r+1)\cdot\,r(r-1)}\right\}}$

$\displaystyle\rm{=\lim_{n\to\infty}\sum^{n}_{r=1}\,\left\{tan^{-1}\left({r}^{2}+r\right)-tan^{-1}\left({r}^{2}-r\right)\right\}}$

This is the telescoping sum, so,

$\displaystyle\rm{=\lim_{n\to\infty}\left\{tan^{-1}\left({n}^{2}+n\right)-tan^{-1}\left(0\right)\right\}}$

$\displaystyle\rm{=\lim_{n\to\infty}\,tan^{-1}\left({n}^{2}+n\right)}$

$\displaystyle\rm{=tan^{-1}\left(\infty\right)}$

$\displaystyle\rm{=\dfrac{\pi}{2}}$