Let $1\leq p\leq\infty$, $\alpha\in\mathbb{R}$ and $I=(0,1]$ with $2\alpha p<1$. For $n\in\mathbb{N}$, Define $f_{n}:I\to\mathbb{R}$ by $f_{n}\left(x\right)=\frac{n}{nx^{\alpha}+1}$ and that $f\left(x\right)=\lim_{n\to\infty}f_{n}\left(x\right)$, $x\in I$. Show that $\left\{ f_{n}\right\} _{n=1}^{\infty}$ converges to $f$ in $L^{p}$, i.e., $\left|\left|f_{n}-f\right|\right|_{p}\to0$.
What I tried:
Proof. WTS that $\lim_{n\to\infty}\int_{0}^{1}\left|f_{n}-f\right|^{p}=0$. $f\left(x\right)=\lim_{n\to\infty}f_{n}\left(x\right)=\lim_{n\to\infty}\frac{n}{nx^{\alpha}+1}=\frac{1}{x^{\alpha}}.$ Now, $$\left|f_{n}-f\right|^{p} =\left|\frac{n}{nx^{\alpha}+1}-\frac{1}{x^{\alpha}}\right|^{p} =\left|\frac{-1}{nx^{2\alpha}+x^{\alpha}}\right|^{p} \leq\frac{1}{nx^{2\alpha p}+x^{\alpha p}} \leq\frac{1}{nx^{2\alpha p}}$$ and this is where i don't know how to proceed. From what I have read, $\frac{1}{nx^2}$ is unbounded. So I'm guessing that $\frac{1}{nx^{2\alpha p}}$ is unbounded. But it says that $2\alpha p<1$, this must have something to do with $\frac{1}{nx^{2\alpha p}}$. Do you have any ideas guys? I will really appreciate it.
We know that $\left|\dfrac{n}{nx^\alpha+1}-\dfrac{1}{x^\alpha}\right|^p=\left|\dfrac{1}{x^\alpha+\dfrac{1}{n}}-\dfrac{1}{x^\alpha}\right|^p\to0$, at $n\to \infty$ for all $x\in(0,1]$. In addition, $\left|\dfrac{1}{x^\alpha+\dfrac{1}{n}}-\dfrac{1}{x^\alpha}\right|^p\leq\dfrac{2^p}{x^{p\alpha}}$ and $\displaystyle\int\limits_0^1\dfrac{2^p}{x^{p\alpha}}dx$ is converges at $p\alpha<1$. So by the Lebesgue theorem of dominant convergence $\displaystyle\int\limits_0^1\left|\dfrac{1}{x^\alpha+\dfrac{1}{n}}-\dfrac{1}{x^\alpha}\right|^pdx\to0.$