Uniform convergence of the series of functions $\sum\limits_{n=1}^{+\infty} {n^n} {\arctan({x}^{n^3})}$

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I have pointwise convergence of the sum of $f_n$ in $(-1,1)$.In fact:

1) if $|x|>1$ $f_n(x)$ doesn't converge to zero

2) if $|x|<1$ $f_n(x) \sim n^n x^{n^3}$ for $n\rightarrow +\infty$ and $\lim\limits_{n\rightarrow +\infty}\sqrt[n]{n^n x^{n^3}}=0$

Is there uniform convergence in $E=(-1,1)$?

$$\sup\limits_E f_n=\max\left(|f_n(-1)|,|f_n(1)|\right)=n^n \frac{\pi}{4}\rightarrow +\infty$$ so there isn't convergence in $E$ if I consider intervals $[a,b]: -1<a<b<1$ there is uniform convergence because

$$\sup\limits_E f_n=\max\left(|f_n(a)|,|f_n(b)|\right)\rightarrow 0$$ for $n\rightarrow +\infty$.