Proving $\sum_{n =1,3,5..}^{\infty }\frac{4k \ \sin^2\left(\frac{n}{k}\right)}{n^2}=\pi$

Proving

$$\sum_{n =1,3,5..}^{\infty }\frac{4k \sin^2\left(\frac{n}{k}\right)}{n^2}=\pi$$ Where $k$ any number greater than $0$

I tried to prove it by using the Fourier series but I couldnt find any form likes the above formula . Any helps. Thanks

Solution 1:

It seems to have escaped attention that this sum may be evaluated using harmonic summation techniques which can be an instructive exercise and usually succeeds on Fourier series.

Suppose we seek to show that $$\sum_{n=1,3,5,\ldots} \frac{\sin^2(n/q)}{n^2} = \frac{\pi}{4q}.$$

As suggested we use $$\sin^2 t = \frac{1-\cos(2t)}{2}$$ to get $$\sum_{n=1,3,5,\ldots} \frac{1}{n^2} \frac{1-\cos(2n/q)}{2}$$ which is $$\frac{1}{2} \sum_{n=1,3,5,\ldots} \frac{1}{n^2} -\frac{1}{2} \sum_{n=1,3,5,\ldots} \frac{1}{n^2} \cos(2n/q).$$

We will be using $$\sum_{n=1,3,5,\ldots} \frac{1}{n^s} = \left(1-\frac{1}{2^s}\right) \zeta(s)$$ which gives for the sum $$\frac{1}{2} \left(1-\frac{1}{2^2}\right) \zeta(2) -\frac{1}{2} \sum_{n=1,3,5,\ldots} \frac{1}{n^2} \cos(2n/q)$$ or $$\frac{\pi^2}{16} -\frac{1}{2} \sum_{n=1,3,5,\ldots} \frac{1}{n^2} \cos(2n/q).$$

Introduce $S(x)%$ given by $$S(x) = \sum_{n=1,3,5,\ldots} \frac{1}{n^2}\cos(nx) = \sum_{k=1}^\infty \frac{1}{(2k-1)^2}\cos((2k-1)x) $$ so that we are interested in $S(2/q).$

The sum term is harmonic and may be evaluated by inverting its Mellin transform.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = \frac{1}{(2k-1)^2}, \quad \mu_k = (2k-1) \quad \text{and} \quad g(x) = \cos(x).$$ We need the Mellin transform $g^*(s)$ of $g(x)$.

Now the Mellin transform of $\cos(x)$ was computed at this MSE link and found to be $$\Gamma(s) \cos(\pi s/2)$$

It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x)$ is given by

$$Q(s) = \Gamma(s)\cos(\pi s/2) \left(1-\frac{1}{2^{s+2}}\right) \zeta(s+2) \\\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{1}{(2k-1)^2} \frac{1}{(2k-1)^s} = \left(1-\frac{1}{2^{s+2}}\right) \zeta(s+2)$$ for $\Re(s) > -1.$

The Mellin inversion integral here is $$\frac{1}{2\pi i} \int_{1/2-i\infty}^{1/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the left for an expansion about zero (the abscissa $\Re(s) = 1/2$ is in the intersection of $\langle -1,\infty\rangle$ and $\langle 0,1\rangle$ from the cosine transform).

The zeros of the cosine term at the negative odd integers cancel the poles of the gamma function at those values. Additional cancelation is gained from the trivial zeros of the zeta function term $\zeta(s+2)$ at the even negative integers $p$ with $p\le -4.$

This leaves just two poles at $s=0$ and $s=1$ and we have $$\mathrm{Res}_{s=0} Q(s)/x^s = \frac{\pi^2}{8} \quad\text{and}\quad \mathrm{Res}_{s=-1} Q(s)/x^s = - \frac{\pi}{4} x$$ and therefore $$S(x) \sim \frac{\pi^2}{8} - \frac{\pi}{4} x.$$ We will see that this is exact for $x\in[0,\pi).$

With $q\ge 1$ we have $2/q\le 2$ and we get for the initial sum the form $$\frac{\pi^2}{16} - \frac{1}{2}\frac{\pi^2}{8} + \frac{1}{2} \frac{\pi}{4} \frac{2}{q} = \frac{\pi}{4q}$$ which is the claim we were trying to prove.

We still need to prove exactness on $[0,\pi)$ to complete the argument.

Put $s= \sigma + it$ with $\sigma \le -3/2$ where we seek to evaluate $$\frac{1}{2\pi i} \int_{-3/2-i\infty}^{-3/2+i\infty} Q(s)/x^s ds$$ by shifting it to the left.

Recall that with $\sigma > 1$ and for $|t|\to\infty$ we have $$|\zeta(\sigma+it)| \in \mathcal{O}(1).$$

Furthermore recall the functional equation of the Riemann Zeta function $$\zeta(1-s) = \frac{2}{2^s\pi^s} \cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s)$$ which we re-parameterize like so $$\zeta(s+2) = 2\times (2\pi)^{s+1} \cos\left(-\frac{\pi (s+1)}{2}\right) \Gamma(-s-1) \zeta(-s-1)$$ which is $$\zeta(s+2) = -2\times (2\pi)^{s+1} \sin(\pi s/2) \frac{\Gamma(1-s)}{s(s+1)} \zeta(-s-1).$$

Substitute this into $Q(s)$ to obtain $$\Gamma(s) \cos(\pi s/2) \left(1-\frac{1}{2^{s+2}}\right) \times -2 \times (2\pi)^{s+1} \sin(\pi s/2) \frac{\Gamma(1-s)}{s(s+1)} \zeta(-s-1).$$

Use the reflection formula for the Gamma function to obtain $$\cos(\pi s/2) \left(1-\frac{1}{2^{s+2}}\right) \times -2 \times (2\pi)^{s+1} \sin(\pi s/2) \times \frac{\pi}{\sin(\pi s)} \frac{1}{s(s+1)} \zeta(-s-1),$$ in other words we have $$Q(s) = -\pi(2\pi)^{s+1} \left(1-\frac{1}{2^{s+2}}\right) \frac{\zeta(-s-1)}{s(s+1)}.$$

There are two components here, call them $Q_1(s)$ and $Q_2(s),$ which are $$-\pi(2\pi)^{s+1} \frac{\zeta(-s-1)}{s(s+1)} \quad\text{and}\quad \pi(2\pi)^{s+1} \frac{1}{2^{s+2}} \frac{\zeta(-s-1)}{s(s+1)}.$$

We evaluate these with $\sigma < -5/2.$ For the first component this implies (with $\sigma = -5/2$ we have $\Re(-s-1) = 3/2$) $$|Q_1(s)/x^s|\sim 2\pi^2 (2\pi)^{\sigma} x^{-\sigma} |t|^{-2}.$$ or $$|Q_1(s)/x^s|\sim 2\pi^2 (x/2/\pi)^{-\sigma} |t|^{-2}.$$

We see from the term in $|t|$ that the integral obviously converges. (This much we knew already.) Moreover, when $x\in(0,2\pi)$ we have $(x/2/\pi)^{-\sigma}\to 0$ as $\sigma\to -\infty.$ The term in $x$ does not depend on the variable $t$ of the integral and may be brought to the front. This means that the contribution from the left side of the rectangular contour that we employ as we shift to the left vanishes in the limit.

For the second component we get $$|Q_2(s)/x^s|\sim \frac{\pi^2}{2} (\pi)^{\sigma} x^{-\sigma} |t|^{-2}.$$ or $$|Q_2(s)/x^s|\sim \frac{\pi^2}{2} (x/\pi)^{-\sigma} |t|^{-2}.$$ This is the same as the first only now we have convergence in $(0,\pi).$

Joining the bounds for $Q_1(s)$ and $Q_2(s)$ we have proved the exactness of the formula for $S(x)$ in the interval $(0,\pi)$ obtained earlier.

As I have mentioned elsewhere there is a theorem hiding here, namely that certain Fourier series can be evaluated by inverting their Mellin transforms which is not terribly surprising and which the reader is invited to state and prove.

Solution 2:

Using (in the interval $(0,\pi)$ ) $$\frac{2\pi x-x^{2}}{8}=\underset{n\geq1}{\sum}\frac{\sin\left(n\frac{x}{2}\right)^{2}}{n^{2}}\,\,(1)$$ we have$$\frac{2\pi x-x^{2}}{8}=\underset{n\geq1}{\sum}\frac{\sin\left(\left(2n-1\right)\frac{x}{2}\right)^{2}}{\left(2n-1\right)^{2}}+\frac{1}{4}\underset{n\geq1}{\sum}\frac{\sin\left(nx\right)^{2}}{n^{2}}$$ and using $(1)$ again we get$$\frac{1}{4}\underset{n\geq1}{\sum}\frac{\sin\left(nx\right)^{2}}{n^{2}}=\frac{\pi x-x^{2}}{8}$$ so$$\frac{\pi x}{8}=\underset{n\geq1}{\sum}\frac{\sin\left(\left(2n-1\right)\frac{x}{2}\right)^{2}}{\left(2n-1\right)^{2}}$$ now put $$x=\frac{2}{k}$$ and we have$$\frac{\pi}{4k}=\underset{n\geq1}{\sum}\frac{\sin\left(\left(2n-1\right)\frac{1}{k}\right)^{2}}{\left(2n-1\right)^{2}}$$ as wanted.