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I need to study the pointwise convergence of

$f_n(x) = [\log(1+x)]^n$

for every $x$ of the domain of the functions.

After i have to prove that the sequence of functions $f_n(x)$ is uniformly convergent to $f$ on the interval $[\frac{1}{2},1]$.

I've shown that in $(-1,\infty)$

$$dom(f_n(x))=\{x:x>-1\}$$

so for $x=0$ $f_n(x)=0 \xrightarrow{} 0$ for n $\rightarrow$ + $\infty$

It remains to show what happen for $x\neq 0$ but i don't find the pointwise convergence. How can i proceed? Thanks in advance for any help.

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@andrew The pointwise limit is clearly $0$ for $x\in(-1+e^{-1},e-1)$. On $[1/2,1]$ you have $$ |\log(x+1)^n-0|=|\log(1+x)|^n<(\log 2)^n\to 0 $$ and this establishes the uniform convergence to zero.

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  • $\begingroup$ how do you understand that the pointwise limit is $0$ for $x \in (-1+e^{-1},e-1)$? $\endgroup$ – andrew Jan 22 at 19:00
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    $\begingroup$ Only for those values you have $\log(1+x)$ is in $(-1,1)$ and increasing powers approach zero. If $x=-1+e^{-1}$ your sequence is $(-1)^n$ which has no limit. If $x=e-1$ it is constant equal to 1, so the limit is $1$ there. Outside the given interval your sequence is $A^n$ with $A\notin[-1,1]$ and is thus has no limit if $A<1$ and the limit is $+\infty$ if $A>1$. $\endgroup$ – GReyes Jan 22 at 19:54
  • $\begingroup$ thank you very much @GReyes $\endgroup$ – andrew Jan 22 at 20:06
  • $\begingroup$ Your first sentence should have an iff. $\endgroup$ – zhw. Jan 22 at 21:08

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