Doubt in calculus

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

@Ajay_2020_1
Ans??

\frac{1}{3}

Getting \dfrac{1}{6}
R u sure of ur ans @Ajay_2020_1

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)

I have got this but confused in last denominator will there be a integration in denominator also ?

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}}

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

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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?

\cot^2(x) = \cot^2(\pi-x)
@pushkar_2020

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