Limit at Infinity $\lim\limits_{m\to\infty}\frac{\sum\limits_{k=1}^m\cot^{2n+1}\left(\frac{k\pi}{2m+1}\right)}{m^{2n+1}}$

Question

How can I prove the following equality? $$\lim_{m\to{\infty}}\frac{\displaystyle\sum_{k=1}^m\cot^{2n+1}\left(\frac{k\pi}{2m+1}\right)}{m^{2n+1}}=\frac{2^{2n+1}\zeta(2n+1)}{\pi^{2n+1}}$$

Answer 1

We can use Paramanand Singh's idea of sandwiching $\cot x$ between $\frac1x$ and something close to that regardless of whether the exponents are odd or even.

The inequality we will use is

$$\frac1x - \frac{x}{2} < \cot x < \frac1x\tag{1}$$

for $0 < x \leqslant \frac{\pi}{2}$. To see $(1)$, note

$$\sin x - x\cos x = \int_0^x y\sin y\,dy > 0$$

for $0 < x \leqslant \pi$, which yields the right hand inequality of $(1)$, and

$$x\cos x - \sin x + \frac{x^2}{2}\sin x = \int_0^x \frac{y^2}{2}\cos y\,dy > 0$$

for $0 < x \leqslant \pi/2$, which yields the left inequality of $(1)$.

So we have

$$\frac{1}{x^{2n+1}} > \cot^{2n+1} x > \frac{1}{x^{2n+1}}\left(1 - \frac{x^2}{2}\right)^{2n+1} > \frac{1}{x^{2n+1}} - \frac{2n+1}{2x^{2n-1}}\tag{2}$$

for $0 < x \leqslant \sqrt{2}$ by the monotonicity of $z\mapsto z^{2n+1}$ and Bernoulli's inequality. Since the two terms on the right hand side of $(2)$ are negative for $\sqrt{2} < x \leqslant \pi/2$, and $\cot x$ is non-negative then, we have

$$\frac{1}{x^{2n+1}} > \cot^{2n+1} x > \frac{1}{x^{2n+1}} - \frac{2n+1}{2x^{2n-1}}\tag{3}$$

for all $0 < x \leqslant \pi/2$. Thus

$$\begin{align} \sum_{k=1}^m \left( \frac{2m+1}{c\cdot k}\right)^{2n+1} &- \left(n+\frac12\right)\sum_{k=1}^m \left(\frac{2m+1}{c\cdot k}\right)^{2n-1}\\ & < \sum_{k=1}^m \cot^{2n+1}\left(\frac{c\cdot k}{2m+1}\right) < \sum_{k=1}^m \left( \frac{2m+1}{c\cdot k}\right)^{2n+1}. \end{align}$$

Now we have

$$ \frac{1}{m^{2n+1}}\sum_{k=1}^m \left( \frac{2m+1}{c\cdot k}\right)^{2n+1} = \frac{(2+1/m)^{2n+1}}{c^{2n+1}}\sum_{k=1}^m \frac{1}{k^{2n+1}} \xrightarrow{m\to\infty} \frac{2^{2n+1}\zeta(2n+1)}{c^{2n+1}} $$

and

$$\frac{n+\frac12}{m^{2n+1}}\sum_{k=1}^m \left(\frac{2m+1}{c\cdot k}\right)^{2n-1} = \frac{\left(n+\frac12\right)(2+1/m)^{2n-1}}{m^2c^{2n-1}}\sum_{k=1}^m \frac{1}{k^{2n-1}} \to 0$$

for $0 < c \leqslant \pi$ and $n > 0$, which yields the desired result.