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For each $n ∈ \Bbb N$, let $f_n : (0, 1) \to \Bbb R$ be defined as`$$f_n(x) = \frac{n}{xn+1}$$ Prove $(f_n )^{∞}_{n=1}$ converges pointwise on $(0,1)$

That is, the $f_n$'s go to zero

I tried finding the limit as n approaches infinity of $\frac{n}{nx+1}$, but that is equal to $\frac{1}{x}$, and given that $x ∈ (0,1)$, this seems contrary to the idea that the sequence converges pointwise. What am I overlooking?

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  • $\begingroup$ You have a sequence, not a series. $\endgroup$ – egreg Dec 5 '18 at 11:04
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$f_n$'s do not go to zero. What you do is correct. We fix $x\in(0,1)$ and then we investigate the limit as $n\to\infty$. We have that, for each fixed $x\in(0,1)$, $f_n(x)\to1/x$ as $n\to\infty$. Hence, $f_n$ converges pointwise to $f$ as $n\to\infty$, where $f$ is defined by $f(x)=1/x$ on $(0,1)$.

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  • $\begingroup$ Thank you! that makes a lotttt of sense $\endgroup$ – user613048 Dec 5 '18 at 11:03

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