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Please see this answer: https://math.stackexchange.com/a/2266612/51407

I want to know why $f_{n}(x)=\frac{x^{\frac{1}{n}}}{( 1+ \frac{x}{n})^{n}} \le \frac{1}{x^{2}}$?

Also, the choice of the domaninating function $g$ is a bit confusing. I can see that if $x \in [M, \infty)$ then $|f_{n}| \le g$, but I don't why if $x \in (0, M)$ then $|f_{n}| \le g$.

EDIT: I have copied the linked proof by user Giovanni below:

Notice that the pointwise limit of the integrands is $e^{-x}$. Moreover, notice that for $M \ge 1$ large enough $$f_n(x) = \frac{x^{\frac{1}{n}}}{(1 + \frac{x}{n})^n} \le \frac{1}{x^2}.$$ Then $$g(x) = \begin{cases}M & \text{if}\ x\in (0,M)\\ \frac{1}{x^2} & \text{if}\ x\in [M,\infty) \end{cases}$$ is integrable and satisfies $|f_n| \le g$ in $(0,\infty)$. By Lebesgue's Dominated Convergence Theorem, $$\lim_n\int_0^{\infty}f_n\,dx = \int_0^{\infty}\lim_nf_n\,dx = \int_0^{\infty}e^{-x}\,dx = 1.$$

Mark
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  • Is it possible to use an alternative option for the dominating function? – Svyatoslav Aug 09 '23 at 01:24
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    On the interval $x\in[0;n]$ $$\frac1{1+\frac xn}=1-\frac xn+\frac {x^2}{n^2}-...=1-\frac xn+\frac {x^2}{n^2}\left(1-\frac xn+\frac {x^2}{n^2}-...\right)=1-\frac xn+\frac xn\frac {\frac xn}{1+\frac xn}$$ $$\leqslant1-\frac xn+\frac xn\cdot\frac12=1-\frac x{2n}$$ $$\frac {x^{\frac1n}}{\big(1+\frac xn)^n}\leqslant x^\frac1n\big(1-\frac x{2n}\big)^n=x^\frac1n e^{n\ln(1-\frac x{2n})}=x^\frac1n e^{n\big(-\frac x{2n}-\frac {x^2}{8n^2}-...\big)}\leqslant n^\frac1ne^{-\frac x2}\leqslant e^\frac1ee^{-\frac x2}$$ as $n^\frac1n$ reaches its maximum at $n=e$ . – Svyatoslav Aug 09 '23 at 01:24
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    Therefore, we can take as a dominating function $f(x)=e^\frac1ee^{-\frac x2}$ on the interval $x\in[0;\infty)$ and consider the sequence of functions $f_n(x)=\frac {x^{\frac1n}}{\big(1+\frac xn)^n}\bf{1}_{[0;n]}$ – Svyatoslav Aug 09 '23 at 01:25
  • @Svyatoslav: I'm interested in an explanation for the original question, but don't mind seeing alternatives either, and I like your answer a lot.

    You should leave your work as an answer also.

     – Mark Aug 09 '23 at 01:47
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    For $x\in[0;1] ,\frac{\frac xn}{1+\frac xn}$ reaches the maximum at $x=n;, \max_{x\in[0;n]}\frac{\frac xn}{1+\frac xn}=\frac12$. Therefore, $$\frac xn\frac{\frac xn}{1+\frac xn}\leqslant\frac xn\cdot\frac12$$ – Svyatoslav Aug 09 '23 at 01:56
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    @Svyatoslav: Yes you're right, thank you! – Mark Aug 09 '23 at 01:57

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