Strange case of uniform convergence of series. Let's consider the following series:
$$\sum_{n = 1}^{\infty}{\left(1 - \cos{\sqrt[3]\frac{x}{n^2}}\right)}$$
for $x$ in the intervals $\delta_1=(0,1)$ and $\delta_2=(1,+\infty)$. It converges uniformly on $\delta_1$ and non-uniformly on $\delta_2$ (this is the answer). I assumed that our series converges and got the same answer.

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*Found derivative and zeros of derivative

*Found supremum of our function $\phi_n(x)$ and said that our series uniformly converges by comparison test (supremum is $1 - \cos{\sqrt[3]\frac{1}{n^2}}$ and if we consider our series converging, then this converges as well)

*Proved that our series does not uniformly converge on $\delta_2$ via negation of Cauchy Criterion of Uniform Convergence

So the only single step left is to prove that given sequence converges. I failed several times.
What I have tried:

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*Ratio test - always gives 1

*Comparison test - I am not sure I know how to find function to compare with in this case. My be there is something about equivalent functions I miss

*I tried to use Maclaurin Series since cosine argument is going to zero, however decomposition to more than one element leading to 1 in Ratio test and decomposition to only $1$ seems strange. By the way I am not sure it is suitable in this kind of tasks at all

I did not try Integral Test, since it seems to be really difficult to integrate this kind of function over $dn$ and I did not try Root Test, since it seems to be useless in our case

Use the identity $1-\cos\theta=2\sin^2(\theta/2)$ to rewrite the series as
$\sum_{n=1}^{\infty}2\sin^2\left(\frac{1}{2}\sqrt[3]\frac{x}{n^2}\right)$.
Now you can apply the comparison test, using the inequality $\sin^2\theta\leq\theta^2$.
