Behavior of function $\sum_{j = n}^\infty \frac{\sin^2((2j-1) \pi x)}{(2j-1)^2}$

Question

For a positive integer $n$, define the function $$ F_n(x) = n^2 \sum_{j = n}^\infty \frac{\sin^2((2j-1) \pi x)}{(2j-1)^2}. $$ I am trying to understand the behavior of $F_n(x)$ in the following sense. For a positive exponent $\alpha$, I would like to compute the limit $$ L_\alpha = \lim_{n \to \infty} F_n(1/n^\alpha). $$

Based on plotting in Mathematica, it appears that $L_\alpha$ has the following behavior: it seems to be $0$ for $\alpha > 2$, infinite for $\alpha < 2$ and some finite number for $\alpha = 2$.

I tried to simplify the summation but I am having difficulty passing to the limit.

Answer 1

Solution.

Fact 1. If $\alpha > 2$, then $\lim_{n \to \infty} F_n(1/n^\alpha) = 0$.

Fact 2. If $1 < \alpha < 2$, then $\lim_{n \to \infty} F_n(1/n^\alpha) = \infty$.

Fact 3. If $0 < \alpha \le 1$, then $\lim_{n \to \infty} F_n(1/n^\alpha) = \infty$.

Fact 4. If $\alpha = 2$, then $\lim_{n \to \infty} F_n(1/n^\alpha) = \frac{\pi^2}{4}$.

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Proof of Fact 1.

Let $N := \lfloor n^\alpha \rfloor$. We have \begin{align*} F_n(1/n^\alpha) &= n^2 \sum_{j = n}^\infty \frac{\sin^2((2j-1) \pi /n^\alpha)}{(2j-1)^2}\\ &= n^2 \sum_{j = n}^{N} \frac{\sin^2((2j-1) \pi /n^\alpha)}{(2j-1)^2} + n^2 \sum_{j = N + 1}^\infty \frac{\sin^2((2j-1) \pi /n^\alpha)}{(2j-1)^2}\\ &\le n^2 \sum_{j = n}^{N} \frac{\left((2j-1) \pi /n^\alpha\right)^2}{(2j-1)^2} + n^2 \sum_{j = N+1}^\infty \frac{1}{(2j-1)^2}\\ &\le n^2 \cdot N \cdot \left( \pi /n^\alpha\right)^2 + n^2 \sum_{j = N+1}^\infty \left(\frac{1}{j-1} - \frac{1}{j}\right)\\ &= n^2 \cdot N \cdot \left( \pi /n^\alpha\right)^2 + n^2 \cdot \frac{1}{N}\\ &\to 0 \end{align*} where we use $\sin^2 u \le u^2$ for all $u > 0$, and $\frac{1}{(2j-1)^2} \le \frac{1}{j^2} \le \frac{1}{(j-1)j} = \frac{1}{j-1} - \frac{1}{j}$ for all $j > 2$.

$\phantom{2}$

Proof of Fact 2.

For sufficiently large $n$, we have $M := \lfloor \frac14n^\alpha\rfloor > n$, and \begin{align*} F_n(1/n^\alpha) &= n^2 \sum_{j = n}^\infty \frac{\sin^2((2j-1) \pi /n^\alpha)}{(2j-1)^2}\\ &\ge n^2 \sum_{j = n}^{M} \frac{\sin^2((2j-1) \pi /n^\alpha)}{(2j-1)^2}\\ &\ge n^2 \sum_{j = n}^{M} \frac{\frac{4}{\pi^2}\left((2j-1) \pi /n^\alpha\right)^2}{(2j-1)^2}\\ &= n^2 (M - n + 1) \frac{4}{\pi^2}\left( \pi /n^\alpha\right)^2\\ &\to \infty \end{align*} where we use $\sin^2 u \ge \frac{4}{\pi^2}u^2$ for all $u \in (0, \pi/2]$, and $(2j-1) \pi /n^\alpha \in (0, \pi/2]$ for all $j \in [n, M]$.

$\phantom{2}$

Proof of Fact 3.

We have \begin{align*} F_n(1/n^\alpha) &= n^2 \sum_{j = n}^\infty \frac{\sin^2((2j-1) \pi /n^\alpha)}{(2j-1)^2}\\ &\ge n^2 \sum_{j = n}^{2n} \frac{\sin^2((2j-1) \pi /n^\alpha)}{(2j-1)^2}\\ &\ge n^2 \sum_{j = n}^{2n} \frac{\sin^2((2j-1) \pi /n^\alpha)}{(2\cdot 2n-1)^2}\\ &= \frac{n^2}{(2\cdot 2n-1)^2} \sum_{j = n}^{2n} \sin^2((2j-1) \pi /n^\alpha)\\ &= \frac{n^2}{(2\cdot 2n-1)^2} \sum_{j = n}^{2n} \frac12\Big(1 - \cos(2(2j-1) \pi /n^\alpha)\Big)\\ &= \frac{n^2(n+1)}{2(2\cdot 2n-1)^2} - \frac{n^2}{2(2\cdot 2n-1)^2} \sum_{j = n}^{2n} \cos(2(2j-1) \pi /n^\alpha). \end{align*}

Let $\beta := 2\pi /n^\alpha$. We have \begin{align*} &\sum_{j = n}^{2n} \cos(2(2j-1) \pi /n^\alpha)\\ ={}& \sum_{j=n}^{2n} \cos ((2j-1)\beta)\\ ={}& \frac{\cos((3n + 1)\beta)\sin((n + 1)\beta)}{\sin \beta} + \cos ((2n-1)\beta) - \cos ((4n + 1)\beta)\\ ={}& \left\{\begin{array}{ll} \cos \frac{2\pi}{n} & \alpha = 1 \\[6pt] O(n^\alpha) & 0 < \alpha < 1. \end{array} \right. \end{align*} Thus, $F_n(1/n^\alpha) \to \infty$.

$\phantom{2}$

Proof of Fact 4.

Let $N := n^2$. We have \begin{align*} F_n(1/n^2) &= n^2 \sum_{j = n}^\infty \frac{\sin^2((2j-1) \pi /n^2)}{(2j-1)^2}\\ &= \frac{\pi^2}{n^2}\sum_{j = n}^\infty \frac{\sin^2((2j-1) \pi /n^2)}{((2j-1) \pi /n^2)^2}\\ &= \frac{\pi^2}{n^2}\sum_{j = 1}^\infty \frac{\sin^2((2j-1) \pi /n^2)}{((2j-1) \pi /n^2)^2} - \frac{\pi^2}{n^2}\sum_{j = 1}^{n-1} \frac{\sin^2((2j-1) \pi /n^2)}{((2j-1) \pi /n^2)^2}\\ &= \frac{\pi}{2} \sum_{j = 1}^\infty \frac{2\pi}{n^2}\cdot \frac{\sin^2((2j-1) \pi /n^2)}{((2j-1) \pi /n^2)^2} - O(1/n)\\ &= \frac{\pi}{2} \sum_{j = 1}^\infty \frac{2\pi}{N}\cdot \frac{\sin^2((2j-1) \pi /N)}{((2j-1) \pi /N)^2} - O(1/n)\\ &\to \frac{\pi}{2} \int_0^\infty \frac{\sin^2 y}{y^2}\mathrm{d} y\\ &= \frac{\pi^2}{4} \end{align*} where we use $\frac{\pi^2}{n^2}\sum_{j = 1}^{n-1} \frac{\sin^2((2j-1) \pi /n^2)}{((2j-1) \pi /n^2)^2} \le \frac{\pi^2}{n^2}\sum_{j = 1}^{n-1} 1 = \frac{\pi^2}{n} = O(1/n)$, and the fact that $\sum_{j = 1}^\infty \frac{2\pi}{N}\cdot \frac{\sin^2((2j-1) \pi /N)}{((2j-1) \pi /N)^2}$ is a Riemann Sum.

Answer 2

This will be a partial answer, since I can't establish what happens when $\alpha<1$, and because I haven't been terribly rigorous in passing from the sum to the integral below. Perhaps someone else can fill in the gaps.

Assume $\alpha \geq 1$. Then: \begin{align} L_\alpha &= \lim_{n\rightarrow\infty} n^2\sum_{j=n}^\infty {\left[\frac{\sin\bigl((2j-1)\pi/n^\alpha\bigr)}{2j-1}\right]}^2\\ &=\lim_{n\rightarrow\infty} \frac{n}{2}\sum_{j=n}^\infty \frac{2}{n}{\left[\frac{\sin\left(\frac{\pi}{n^{\alpha-1}}\,\frac{2j-1}{n}\right)}{\frac{2j-1}{n}}\right]}^2 \qquad\text{Riemann sum with $x = (2j-1)/n$}\\ &\rightarrow\lim_{n\rightarrow\infty} \frac{n}{2}\int_2^\infty {\left[\frac{\sin\left(\frac{\pi}{n^{\alpha-1}}\,x\right)}{x}\right]}^2 dx\qquad\;\;\,\text{Change of variables $x = y n^{\alpha-1}$}\\ &=\lim_{n\rightarrow\infty} \frac{n^{2-\alpha}}{2}\int_{2/n^{\alpha-1}}^\infty {\left[\frac{\sin\left(\pi\,y\right)}{y}\right]}^2 dy \qquad\;\;\;(1)\\ \end{align} There are now two cases to consider:

$\bullet\;\alpha>1$

Here (1) becomes \begin{align} L_\alpha &= \lim_{n\rightarrow\infty} \frac{n^{2-\alpha}}{2}\int_0^\infty {\left[\frac{\sin\left(\pi\,y\right)}{y}\right]}^2 dy\\ &=\lim_{n\rightarrow\infty} n^{2-\alpha}\;\frac{\pi^2}{4} \end{align} Clearly this diverges for $1<\alpha<2$, goes to zero for $\alpha>2$, and goes to $\pi^2/4$ for $\alpha=2$.

$\bullet\;\alpha=1$

Here (1) becomes \begin{equation} L_\alpha = \lim_{n\rightarrow\infty} \frac{n}{2}\int_2^\infty {\left[\frac{\sin\left(\pi\,y\right)}{y}\right]}^2 dy \rightarrow \infty \end{equation}

I don't know what happens when $\alpha<1$ because in that case the Riemann sum above can't be simply written as the integral that follows it.

Answer 3

Let me verify the case $\alpha = 2$.

Using the identity $\sin^2(x) = \frac{1}{2}(1 - \cos(2x))$ and the well known sum $\sum_{j = 1}^\infty \frac{1}{(2j-1)^2} = \frac{\pi^2}{8}$, write $$ \begin{align*} n^{-2}F_n(x) &= \sum_{j = n}^\infty \frac{1}{2(2j-1)^2} - \sum_{j = n}^\infty\frac{\cos(2 \pi (2j - 1) x)}{2(2j-1)^2} \\ &= \frac{\pi^2}{16} -\sum_{j=1}^{n-1} \frac{1}{2(2j-1)^2} -\sum_{j = n}^\infty\frac{\cos(2 \pi (2j - 1) x)}{2(2j-1)^2} \\ \end{align*} $$ Now recall the Fourier series expansion of $|x|$ for $x \in (-\frac{1}{2}, \frac{1}{2})$ : $$ |x| = \frac{1}{4} - \sum_{j=1}^\infty \frac{2\cos(2\pi(2j-1)x)}{\pi^2(2j -1)^2} $$ Substituting this in the above, we obtain $$ \begin{align*} n^{-2}F_n(x) &= \frac{\pi^2}{16} -\sum_{j=1}^{n-1} \frac{1}{2(2j-1)^2}+ \frac{\pi^2}{4}|x| -\frac{\pi^2}{16}+\sum_{j=1}^{n-1} \frac{\cos(2\pi(2j-1)x)}{2(2j -1)^2} \\ &= \frac{\pi^2}{4}|x| -\sum_{j=1}^{n-1} \frac{1 -\cos(2\pi(2j-1)x)}{2(2j -1)^2} \\ \end{align*} $$ Thus $$ \begin{align*} F_n(n^{-2}) &= \frac{\pi^2}{4} -n^2\sum_{j=1}^{n-1} \frac{1 -\cos(2\pi(2j-1) n^{-2})}{2(2j -1)^2}\\ &= \frac{\pi^2}{4} -n^2\sum_{j=1}^{2n-3} \frac{\sin^2(\pi j n^{-2})}{j^2}. \end{align*} $$

Denote $G(x):=\frac{\sin^2(\pi x)}{x^2}$, and notice that the last sum is $\frac{1}{n^2} \sum_{j=1}^{2n-3} G(j/n^2)$.

Since $G$ is bounded, $$ \limsup_{n\to \infty}\frac{1}{n^2}\sum_{j= 1}^{2n-3} G(j/n^2) \le C\limsup_{n\to \infty} \frac{2n-3}{n^2} = 0, $$ Thus indeed $$ \lim_{n\to\infty} F_n(n^{-2}) = \frac{\pi^2}{4}. $$ Remark. We have the formula $$ F_n(n^{-\alpha}) = \frac{\pi^2}{4}n^{2-\alpha} - \frac{1}{n^{(2-2\alpha)}} \sum_{j=1}^{2n-3} G(j/n^{\alpha}) \le \frac{\pi^2}{4}n^{2-\alpha} , $$ from which the cases $\alpha > 2$, and $0 < \alpha \le 1$ follow easily.