prove function series $f_n(x)=(\sin x)^{1/n} $ doesn' converge uniformly

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I'm trying to prove that the following function series doesn't uniformaly converge on $(0,\pi/2]$ but I'm not sure I'm doing it right. I would really appreciate if someone could guide me and explain my mistakes.

my idea: (I marked with yellow marker the main points I think are faulty)

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at my 1st main question the segments are $I=[\alpha,\pi/2]$ and $J=(0,\pi/2]$

Thank you!

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If the convergence was uniform, then you would also have uniform convergence on $[0,\frac{\pi}{2}]$.

Now, the simple limit on $[0, \frac{\pi}{2}]$ of the sequence $(f_n)$ is the function $f$ such that $f(x) = 1$ for all $x \in (0, \frac{\pi}{2}]$ and $f(0) = 0$, so the simple limit is not continuous. Therefore the convergence cannot be uniform.