Behaviour of a sequence - how to fix this faulty proof? Today I am resurrecting an interesting past question. The original post sounded like

$\{a_k\}_{k\geq 1}$ is a sequence of positive real numbers and
  $$\left|\sum_{k\geq 1}\frac{\sin(a_k x)}{k^2}\right|\leq\left|\tan(x)\right| $$
  holds for any $x\in(-1,1)$. Show that $a_k=o(k^2)$.

My faulty solution was to divide both sides of the given inequality by $x$ and to consider the limit as $x\to 0^+$,
$$ \left|\sum_{k\geq 1}\frac{a_k}{k^2}\right|\leq 1 $$
in order to deduce that $a_k=o(k^2)$. The HUGE issue is that the exchange of $\lim_{x\to 0^+}$ and $\sum_{k\geq 1}$ is completely unjustified, and I am struggling in showing that $\sum_{k\geq 1}\frac{\text{sinc}(a_k x)}{k^2}$ is uniformly convergent. I am not even sure the hypothesis ensure such uniform convergence. So, for short,

How to fix this faulty proof?

Is there some slick application of the dominated convergence theorem or of the Banach-Steinhaus theorem?
 A: Let us discuss if this variant is sound. It is safe to state that for any $x\in(0,1)$
$$ \left|\sum_{k\geq 1}\frac{\sin(a_k x)}{x k^2}\right|\leq\left|\frac{\tan x}{x}\right| $$
as well as
$$\lim_{x\to 0^+}\left|\sum_{k\geq 1}\frac{\sin(a_k x)}{x k^2}\right|=\left|\lim_{x\to 0^+}\sum_{k\geq 1}\frac{\sin(a_k x)}{x k^2}\right|\leq\lim_{x\to 0^+}\left|\frac{\tan x}{x}\right|=1.$$
Now $\sum_{k\geq 1}\frac{\sin(a_k x)}{x k^2}=f(x)$ is a continuous function over any compact subset of $\mathbb{R}^+$ and $|f(x)|\leq\frac{\zeta(2)}{x}$. We are allowed to compute $\lim_{x\to 0^+}f(x)$ through a convolution with an approximate identity, i.e. through
$$ \lim_{x\to 0^+}f(x)=\lim_{M\to +\infty}\int_{0}^{+\infty}M e^{-Mx} f(x)\,dx. $$
This gives
$$ \lim_{x\to 0^+}\sum_{k\geq 1}\frac{\sin(a_k x)}{xk^2} =\lim_{M\to +\infty}M\int_{0}^{+\infty}\sum_{k\geq 1}\frac{\sin(a_k x)e^{-Mx}}{x k^2}\,dx=\lim_{M\to +\infty}\sum_{k\geq 1}\frac{M\arctan\left(\frac{a_k}{M}\right)}{k^2}$$
and $\sum_{k\geq 1}\frac{M\arctan\left(\frac{a_k}{M}\right)}{k^2}$ is an increasing function of the $M$-variable (as a sum of non-negative and increasing functions). This gives
$$ \lim_{x\to 0^+}\sum_{k\geq 1}\frac{\sin(a_k x)}{x k^2} =\sup_{M}\sum_{k\geq 1}\frac{M\arctan\left(\frac{a_k}{M}\right)}{k^2}.$$
Let us assume that for some positive constant $C$ the inequality $a_k\geq Ck^2$ holds infinite times, in particular for $a_{k_1},a_{k_2},\ldots,a_{k_H}$, and let us consider $M=2a_{k_H}$. Since $\arctan(x)\geq\frac{9x}{10}$ over $\left[0,\frac{1}{2}\right]$,
$$ \sup_{M}\sum_{k\geq 1}\frac{M\arctan\left(\frac{a_k}{M}\right)}{k^2}\geq \sum_{k\geq 1}\frac{M\arctan\left(\frac{a_k}{M}\right)}{k^2}\geq \sum_{h=1 }^{H}\frac{M\arctan\left(\frac{a_{k_h}}{M}\right)}{k_h^2}\geq\frac{9}{10}\sum_{h=1}^{H}\frac{a_{k_h}}{k_h^2}\geq\frac{9}{10}CH.$$
By taking $H=\left\lfloor\frac{2}{C}\right\rfloor$ we deduce
$$ \lim_{x\to 0^+}\sum_{k\geq 1}\frac{\sin(a_k x)}{x k^2}=\sup_{M}\sum_{k\geq 1}\frac{M\arctan\left(\frac{a_k}{M}\right)}{k^2}\geq \frac{3}{2} $$
which contradicts the fact that the LHS is bounded by $1$ in absolute value. This proves that $a_k=o(k^2)$ as wanted.
