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I want to find its pointwise convergent sequence, if any.

$f_n$ itself converges pointwise everywhere, and there will be a pointwise convergent sequence of odd and even, so at least two.

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  • $\begingroup$ Can you specify what you mean by "pointwise convergent sequence"? Sequence on $n$ for fixed $x$? Sequence on $x$ for fixed $n$? Sequences on $n,x$? $\endgroup$ Dec 21, 2016 at 9:11

1 Answer 1

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For $x\in[0,1)$, $f_n(x)=(-1)^nx^n\to0$ since $(-1)^n$ is bounded by $1$ and $x^n\to0$.

For $x=1$, $f_{2n}(x)=1$ and $f_{2n-1}(x)=-1$ so you do not have convergence at $x=1$.

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  • $\begingroup$ ohh yes $f_n$ it self is not pointwise convergent everywhere as on $x=1$ $\endgroup$
    – blabla
    Dec 21, 2016 at 9:16

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