Is $\lim_{m \to 0+} \sum\limits_{k=1}^{\infty} k^{\left(\frac{1}{1+m}\right)} \sin\left(\frac{\pi x}{k^2}\right)$ ~= $K x$?

Question

Sorry for posting a very similar question as my previous one and also this question. I plotted the following with a combination of wolfram-alpha and excel. It looks almost like a straight line passing through origin, but it definitely flattens as $x$ increases. I don't know how accurately wolfram-alpha calculates, but it does report that the sum diverges for $m = 0$, but converges for $m > 0$. Whether there is any simple curve that can be fit to this expression or not, one interesting thing I noticed is that as $m$ approaches zero the curve is monotonically increasing (whereas with $m$ increasing, I can see some local maxima and minima).

$$f(x) = \lim_{m \to 0+} \sum\limits_{k=1}^{\infty} k^{\left(\frac{1}{1+m}\right)} \sin\left(\frac{\pi x}{k^2}\right)$$

Does this function converge to a polynomial?

Answer 1

For each fixed $x \neq 0$, write

$$ \sin\left(\frac{\pi x}{k^2}\right) = \frac{\pi x}{k^2} + \mathcal{O}\left(\frac{1}{k^6}\right) $$

where the implicit bound for the Big-Oh term depends only on $x$. Then it follows that

$$ \sum_{k=1}^{\infty} k^{\frac{1}{1+m}} \sin\left(\frac{\pi x}{k^2}\right) = \pi x \zeta\left(\frac{1+2m}{1+m}\right)+\mathcal{O}(1) $$

as $m \to 0^+$. This shows that for each $x \neq 0$,

$$ \sum_{k=1}^{\infty} k^{\frac{1}{1+m}} \sin\left(\frac{\pi x}{k^2}\right) \sim \frac{\pi x}{2m} \quad \text{as } m \to 0^+. $$

Consequently $f(x)$ does not converge unless $x = 0$.

Answer 2

In the same spirit as Sangchul Lee's answer, for all $x$ we have $$\sin (x) = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1} $$ which make $$ \sin\left(\frac{\pi x}{k^2}\right)= \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!}\left(\frac{\pi x}{k^2}\right)^{2n+1}$$ which makes $$\sum^{\infty}_{k=1}k^{\frac{1}{1+m}} \sin\left(\frac{\pi x}{k^2}\right)=\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!}(\pi x)^{2n+1}\zeta \left(4 n+2-\frac{1}{m+1}\right)$$ Now expanding as series around $m=0$ terms $$u_n=\zeta \left(4 n+2-\frac{1}{m+1}\right)$$ we have $$u_0=\frac{1}{m}+(1+\gamma )-m \gamma _1+O\left(m^2\right)$$ where appear Euler's and Stieltjes constants. For any other term $$u_n=\zeta (4n+1)+ \zeta '(4n+1)\,m+O\left(m^2\right)$$ So, as Sangchul Lee explained, when $m\to 0$, the problem is really related to the first term.