find \\ \displaystyle \lim_{n \to \infty} \dfrac{\sum_{r = 1}^{n-1}\cot^2 \left(\frac{r \pi}{n} \right )}{n^2}

# Doubt in calculus

\frac{1}{3}

Yes

as n \rightarrow \infty

cot^2 {\theta} \approx {\theta}^2

by squeeze theoram u will get result

source of beautiful problem @Ajay_2020_1

now i got some idea

these are roots of equation tan(rx)=0

expand tan(rx) by fornula u will get exact ans

r=n i mean tan(nx)

second line is wrong cant take log in that way

by the way u will not get any result by \textbf{rieman sum}

Ok thank you for telling

to calculate the sum first we will convert the summation into half i.e

\\ \displaystyle \\ \displaystyle \sum_{r = 1}^{n-1}\cot^2 \left(\frac{r \pi}{n} \right ) = \sum_{r = 1}^{n-1}\cot^2 \left(\frac{{(n - r)} \pi}{n} \right ) so

\displaystyle \ \sum_{r = 1}^{n-1}\cot^2 \left(\frac{r \pi}{n} \right ) = 2\sum_{r = 1}^{[\frac{n}{2}]}\cot^2 \left(\frac{r \pi}{n} \right )

We did this just to avoid the cot being negative and the idea is to apply squeeze theorem in the summation.

so for squeeze theorem

```
we will consider the inequality
```

\displaystyle \frac{1}{x^2} +\frac{16}{{\pi}^4}-\frac{8}{\pi^3} < \cot^2(x) < \frac{1}{x^2} 'can be proved using aod'

and now sum up by putting x = \frac{r\pi}{n}

denote

\sum_{r = 1}^{[\frac{n}{2}]}\cot^2 \left(\frac{r \pi}{n} \right ) by \mathfrak{J}

so \displaystyle \frac{n^2}{\pi^2}\sum_{r=1}^{r = \left[\frac{n}{2} \right]}\frac{1}{r^2}+(\frac{16}{{\pi}^4}-\frac{8}{\pi^3}) [\frac{n}{2}] < \mathfrak{J} < \frac{n^2}{\pi^2}\sum_{r=1}^{r = \left[\frac{n}{2} \right]}\frac{1}{r^2}

now dividing by n^2 and taking limit n \to \infty

we get \mathfrak{J} = \frac{1}{6}

which gives you the answer as \boxed{\frac{1}{3}}

here i have used the fact that \displaystyle \zeta(2) = \frac{\pi^2}{6}

another solution uses complex numbers given in the solution

See same approach as mine

Another u can prove by induction

Another by using vieta relation roots u will get same approach

bro didn't understand..floor function raise aaya?