For which $p \in [1, \infty]$ the sequence is Cauchy

Let $f_n (x) = \frac{n}{1+n\sqrt{x}}$, for $x \in (0,1)$. I am asked to find for which $p \in [1, \infty]$ $(f_n)_n$ is a Cauchy sequence with respect to $L^p$ norm.

First of all, I observed that $(L^p(0,1), ||\cdot||_{L^p})$ is a Banach space for all $1 \le p \le \infty$. Thus, $(f_n)_n$ is a Cauchy sequence $\Leftrightarrow$ $(f_n)_n$ is convergent in $L^p(0,1)$. Moreover, since $\lim_{n \rightarrow \infty} f_n (x) = f(x) := \frac{1}{\sqrt{x}} \in L^p(0,1), 1 \le p < 2$, for all $x \in (0,1)$, the strong limit - if it exists - must coincide with $f$.

The case $p=1$ is easy, as it is possible to show that $||f_n - f||_{L^1} = \int_{0}^{1} \frac{1}{\sqrt{x} + nx} dx = \frac{2 log(n+1)}{n} \rightarrow 0$ as $n$ approaches infinity. Hence, the sequence being convergent in $L^p(0,1)$ is Cauchy.

For the case $1 < p < \infty$ (and similarly for $p=\infty$), I proceeded in a similar way. However, in this case I end up with an integral that does not converge for all $n \in \mathbb{N}$, that is $||f_n - f||_{L^1} = \int_{0}^{1} \frac{1}{\sqrt{x} + nx}^p dx = \left.\frac{1}{n} \frac{(t + n t^2)^{1-p}}{p-1}\,\right|_{0}^{1} \rightarrow \infty$ as $x \rightarrow 0$.

Is it correct to state that in this case the sequence does not converge (and thus it can't be a Cauchy sequence)? My doubt here concerns the fact that I am not able to estimate this integral with something that is convergent as $x$ approaches $0$ and hence I cannot take the limit for $n \rightarrow \infty$ and conclude.

If anyone could give me a feedback on my reasoning or suggest a more straight-forward way to solve the problem, it would be greatly appreciated.

Convergence in any $L^{p}$ implies almost everywhere convergence for a subsequence. So the sequence can only converge to $f(x)=\frac 1 {\sqrt x}$. For $1 \le p < 2$ we have $f \in L^{p}$ and $0 \leq f_n(x) \leq \frac 1 {\sqrt x}$ and so $\int |f_n(x)-f(x)|^{p} \to 0$ by DCT. [Use the dominating function $2^{p+1} f^{p}$]. For $p \ge 2$ the function $f$ does not belong to $L^{p}$ and this implies that $(f_n)$ cannot converge in $L^{p}$.