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I want to study the pointwise convergence of $f_n(x) = (x+1)\arctan(x^n)$ on $R$ but I have trouble establishing pointwise convergence on the interval $I = [-\infty, -1)$. My reasoning is the following:

When $x\in I$, $x^n$ diverges, hence, $\arctan(x^n)$ is also divergent. Since $\arctan(x^n)$ diverges as $n \to \infty$, $f_n(x)$ is also divergent.

Is my reasoning correct or does this count as a rigorous proof at all? Thanks.

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It's not the fact that $x^n$ diverges that makes $\arctan(x^n)$ diverge, it's the fact that $x^n$ is alternately $> 1$ and $< -1$.

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    $\begingroup$ So can I say that if $n = 2k$, $f_n(x) \to \frac{\pi}{2} (x+1)$ and when $n = 2k + 1$, $f_n(x) \to - \frac{\pi}{2} (x+1)$? $\endgroup$
    – Alderson
    Commented Apr 20, 2017 at 19:23

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