Consider for $x>0$
$$f(x) = \frac{\sum_{n=1}^{\infty} \sin^2(x/n)}{x}$$
I was fascinated by the behaviour of this function.
It is easy to show that
$$\lim_{x \to +0} \frac{f(x)}{x} = \zeta(2) = \frac{\pi^2}{6}$$
It seems we get for all $x>1$
$$f(1) \leq f(x) \leq \zeta(2)$$
and for all sufficiently large $x > d$ (for some $d$) :
$$ C(d) \leq f(x) $$
for some nonzero $C(d)$ larger than $f(1)$.
But also the remarkable
$$f(y) = \zeta(2)$$
for some mysterious $y$ around $4.56...$
I guess $C$ and $y$ have closed forms too.
Let $x>5$, then I wonder about the possible existance of
$$ \lim \sup f(x) = A $$
and is $A = \zeta(2)$ ?
Similarly
Let $x>5$, then I wonder about the possible existance of
$$ \lim \inf f(x) = B $$
and is $C(d) < B$ for any choice of $d$ ?
How often do we get a local maximum in a given real interval ?
The function $f(x)$ converges pretty slowly, what makes computation harder.
It would be nice to have some plots.
This seems more like a general request for information than a specific question, so I'll give some analysis, including
Also, here is a Desmos graph with a some of the mentioned formulas: https://www.desmos.com/calculator/mlgjz3cu0b
First, note that the term in the sum is positive, so for positive $x$, $f(x)$ is greater than its partial sums. Each partial sum is a lower bound for $f$. We can find an upper bound like this:
\begin{align} f(x) - \sum_{n = 1}^N \frac{\sin^2(x/n)}x &= \sum_{n = N + 1}^\infty \frac{\sin^2(x/n)}x\\ &\leq \sum_{n = N + 1}^\infty \frac{x^2/n^2}x\\ &= x\sum_{n = N + 1}^\infty \frac1{n^2}\\ &\leq x\int_N^\infty \frac1{t^2}dt\\ &= \frac xN. \end{align}
Therefore, $$ \sum_{n = 1}^N \frac{\sin^2(x/n)}x + \frac xN $$ is an upper bound for $f(x)$. Conveniently, this upper bound converges faster than the original expression, so you can graph it to see a more accurate picture of the function.
Now, I'll show that $f(x) \to \pi/2$ as $x \to \infty$. For every natural number $N$, let $$ f_N(x) = \sum_{n = 1}^{\lceil Nx\rceil} \frac{\sin^2(x/n)}x. $$ Clearly, $f_N(x)$ converges pointwise to $f(x)$ as $N \to \infty$. Applying the upper bound we just found, we can see that $$ |f(x) - f_N(x)| = f(x) - \sum_{n = 1}^{\lceil Nx\rceil} \frac{\sin^2(x/n)}x < \frac x{Nx} = \frac1N, $$ which implies that the convergence is uniform. This means that we can interchange limits of $x$ and $N$, so $$ \lim_{x \to \infty} f(x) = \lim_{x \to \infty}\left(\lim_{N \to \infty} f_N(x)\right) = \lim_{N \to \infty}\left(\lim_{x \to \infty} f_N(x)\right). $$
Now, we will see that $f_N(x)$ is a Riemann sum of $\sin^2(1/t)$, which will allow us to convert it into an integral. Consider the partition $p = \{\frac1x, \frac2x, \frac3x, \dots, \frac{\lceil Nx\rceil + 1}x\}$. Then $$ f_N(x) = \sum_{n = 1}^{\lceil Nx\rceil} \sin^2\left(\frac1{t^*_n}\right) \Delta t_n, $$ where $t^*_n = n/x$ and $\Delta t_n = 1/x$. This is a Riemann sum of $\sin^2(1/t)$ over $p$. Since $\sin^2(1/t)$ is Riemann-integrable for positive numbers and $\|p\| = 1/x \to 0$ as $x \to \infty$, we have $$ \lim_{x \to \infty} f_N(x) = \int_0^N \sin^2\left(\frac1t\right)dt. $$ This means that $$ \lim_{N \to \infty}\left(\lim_{x \to \infty} f_N(x)\right) = \int_0^\infty \sin^2\left(\frac1t\right)dt. $$
Substituting $u = 1/t$ transforms the integral to $$ \int_0^\infty \frac{\sin^2(u)}{u^2}du, $$ which is proved to equal $\pi/2$ by the answers to this question. Therefore $$ \lim_{x \to \infty} f(x) = \frac\pi2. $$
To find the power series expansion, note that \begin{align} f(x) = \sum_{n = 1}^\infty \frac{\sin^2(x/n)}x &= \sum_{n = 1}^\infty \frac{1 - \cos(2x/n)}{2x}\\ &= \sum_{n = 1}^\infty \frac{1 - \left(1 - \frac1{2!}\left(\frac{2x}n\right)^2 + \frac1{4!}\left(\frac{2x}n\right)^4 - \frac1{6!}\left(\frac{2x}n\right)^6 + \cdots\right)}{2x}\\ &= \sum_{n = 1}^\infty \frac{\sum_{k = 1}^{\infty}(-1)^{k-1} \frac1{(2k)!}\left( \frac{2x}n \right)^{2k}}{2x}\\ &=\sum_{n=1}^\infty \sum_{k=1}^\infty (-1)^{k-1} \frac1{n^{2k}} \cdot \frac{(2x)^{2k-1}}{(2k)!}. \end{align}
Because Taylor series converges absolutely, we know that the inner sum converges absolutely. And since the inner sum is $O(1/n^2)$, the double sum must converge absolutely. Therefore, we can swap the sums, or equivalently, swap $n$ and $k$.
\begin{align} &=\sum_{n=1}^\infty \sum_{k=1}^\infty (-1)^{n-1} \frac1{k^{2n}} \cdot \frac{(2x)^{2n-1}}{(2n)!}\\ &=\sum_{n=1}^\infty (-1)^{n-1} \frac{(2x)^{2n-1}}{(2n)!} \sum_{k=1}^\infty \frac1{k^{2n}}\\ f(x) &= \sum_{n=1}^\infty (-1)^{n-1} \frac{\zeta(2n)(2x)^{2n-1}}{(2n)!}. \end{align}