Answered: "What is the value of: limn→∞∑r=1n−1cot2(rπn)n2"

Branden Valentine

Answered question

2022-01-27

What is the value of:

lim_{n \to \infty} \sum_{r = 1}^{n - 1} \frac{\cot^{2} (r \frac{π}{n})}{n^{2}}

Answer & Explanation

ebbonxah

Beginner2022-01-28Added 15 answers

It would be naïve and incorrect to proceed as follows
$\sum_{k = 1}^{n - 1} \frac{\cot^{2} (π \frac{k}{n})}{n^{2}} \underset{WRONG!}{\underset{⏟}{\approx}} \frac{1}{n} \int_{\frac{1}{n}}^{1 - \frac{1}{n}} \cot^{2} (π x) d x$
$= \frac{1}{n} (- x - \frac{1}{π} \cot (π x)) ∣_{\frac{1}{n}}^{1 - \frac{1}{n}}$
$= \frac{2}{n^{2}} - \frac{1}{n} + \frac{2}{n π} \cot (\frac{π}{n})$
$\to \frac{2}{π^{2}}$
Instead, we use $\cot^{2} (x) = \csc^{2} (x) - 1$ to write
$\sum_{k = 1}^{n - 1} \frac{\cot^{2} (π \frac{k}{n})}{n^{2}} = \frac{1}{n} - \frac{1}{n^{2}} + \sum_{k = 1}^{n - 1} \frac{1}{n^{2}, \sin^{2} (π \frac{k}{n})}$
$= \frac{1}{n} - \frac{1}{n^{2}} + 2 \sum_{k = 1}^{| n / 2 | - 1} \frac{1}{n^{2} \sin^{2} (π k / n)}$
Next, we note that for $\frac{π}{2} > x > 0, {(x - {\frac{16}{x}}^{3})}^{2} \leq \sin^{2} (x) \leq x^{2}$ Hence, we have
$\frac{2}{π^{2}} \sum_{k = 1}^{| \frac{n}{2} | - 1} \frac{1}{k^{2}} \leq \frac{2}{n^{2}} \sum_{k = 1}^{| \frac{n}{2} | - 1} \frac{1}{n^{2}, \sin^{2} (π \frac{k}{n})} \leq \frac{2}{π^{2}}$

coolbananas03ok Beginner2022-01-29Added 20 answers

By applying Vieta's formulas to Chebyshev polynomials of the second kind we have

\sum_{r = 1}^{n - 1} \cot^{2} (\frac{π r}{n}) = \frac{(n - 1) (n - 2)}{3}

(compare Cauchy's proof of

ζ (2) = \frac{π^{2}}{6}

in his Cours d'Analyse) hence the wanted limit is clearly

\frac{1}{3}

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Precalculus

Didn't find what you were looking for?