Showing that $1/\sqrt{x}\in L^1$ by finding Cauchy sequence that converges to it I am struggling with a homework example and I hope you could give me a hint on it. The task is as follows:

Show that $f(x) = \frac{1}{\sqrt{x}}$ is an element of $L^1 \big( (0,1) \big)$ by finding a Cauchy sequence $f_n$ in $\big( C^{\infty}([0,1]) \, , \; \| \cdot \|_{L^1(0,1)} \big)$ that  converges to $f$.

Now, as I understand it, in order to show that $f \in L^1(0,1)$, one uses completeness of the space $\big( C^{\infty}([0,1]) \, , \; || \cdot ||_{L^1(0,1)} \big)$. So, if one finds a Cauchy sequence in that space that converges to $f$, then $f$ must be in $\big( C^{\infty}([0,1]) \, , \; || \cdot ||_{L^1(0,1)} \big)$ as well, which means that the norm of $f$ is finite and therefore $f \in L^1(0,1)$. Is this interpretation correct?
If so, I am struggling to find such a sequence. Finding a sequence that is an element of $C^{\infty}([0,1])$ and that converges to $f$ is not that hard, e.g.
$$f_n = \sqrt{\frac{1}{x + \tfrac{1}{n}}}$$
However, I am having difficulty finding a sequence that is Cauchy. Are there any hints or tips you could give me on finding such a Cauchy sequence?

EDIT 1:
It has been mentioned that the above sequence $f_n$ itself is Cauchy. I tried proving this, but failed. My approach was
$$|| f_n - f_m ||_{L^1(0,1)} = \int_0^1 | f_n(x) - f_m(x) | dx \leq \int_0^1 |f_n(x)| dx + \int_0^1 |f_m(x)| dx$$
Consider
$$\int_0^1 f_n(x) dx = \int_0^1 \frac{1}{\sqrt{x + \tfrac{1}{n}}}dx = \sqrt{n} \int_0^1 \frac{1}{\sqrt{nx + 1}}dx$$
How to proceed, or is this approach flawed anyway?
 A: Reiterating my comment: I find the question rather odd. Its one thing to find such a Cauchy sequence. But there are certainly much easier ways to show that $1/\sqrt x$
is integrable on $[0,1]$.
Your computation cannot work because it does not try to take advantage of the cancellation you get when you subtract two similar things in $f_n - f_m$. The best bound you can get once you end up at $\int |f_n| +\int |f_m|$ is essentially 2$\int |f|$ (since $f_n\to f$ in $L^1$ implies $\|f_n\|_{L^1} \to \|f\|_{L^1}$.)
It is not too hard to prove that your particular choice of $f_n$  is Cauchy, but it is awkward to work without first proving $1/\sqrt x$ is integrable. And in fact it is simpler to just prove that $f_n$ straight up converges to $f$ with dominated convergence theorem: $f_n-f\to 0$ a.e. and $|f_n - f| \le |f|+|f_n| \le 2|f| \in L^1((0,1)) $, so $\int |f_n-f| \to 0$, QED.
Here's how to do it anyway. The sketch if you wish to try further yourself: you can split estimating $\int_0^1|f_n-f_m|$ into a region around the "problem point" $\int_0^c|f_n-f_m|$, $c\in(0,1)$, and a region away from it $\int_c^1|f_n-f_m|$. We control the first integral by using $c\ll1$ (Here we have to somehow avoid appealing to $\frac1{\sqrt{x}}\in L^1$, at least directly). In the second region, we are dealing with the convergence of continuous functions, and we should know how this part goes. Putting the two estimates together by choosing $c$ well gives the result.
First note that for a $c>0$ to be chosen, if say $n>m$,
$$\int_0^c |f_m-f_n| = \int_0^c f_n-f_m  \le \int_0^c f_n = \int_{1/n}^{c+1/n} \frac{dx}{\sqrt x}  = 2\sqrt {c+1/n}-2\sqrt{1/n} \le  2\sqrt c.$$(This is just an awkward rearrangement of the proof that $\frac1{\sqrt x}\in L^1$. We used that $\sqrt{a+b}\le\sqrt{a}+\sqrt b$, though we don't need to...)  Bounding the $[c,1]$ part with triangle inequality, we therefore have (with $\|v\|_{L^\infty(A)} := \sup_{x\in A} |v(x)|$)
$$ \|f_n - f_m\|_{L^1([0,1])} \le 2\sqrt c + (1-c)\|f_n-f_m\|_{L^\infty([c,1])}. $$
On the compact interval $[c,1]$, $f_n\to f$ pointwise and are continuous; hence $f_n\to f$ uniformly, and so $f_n$ is Cauchy in $L^\infty([c,1])$.
Now, for any fixed $\epsilon>0$, setting $\sqrt c=\epsilon$, then taking $N\gg1$ so that $n,m>N$ implies $\|f_n-f_m\|_{L^\infty([c,1])}<\epsilon$, the above gives $\|f_n - f_m\|_{L^1} \le 3\epsilon$, QED.
Yet another way you could proceed- first find a $C^0([0,1])$ sequence $g_n$ converging to $f$ in $L^1([0,1])$. One candidate could be $f_n(x) = \min(n,f)$. Once you show $g_n\to f$, you can use Weierstrass approximation to find polynomials $f_n$ (hence smooth) so close to $g_n$, that $f_n\to f$ as well.
A: Explore $x_{i+1} = (x_i + x/x_i)/2$, $0 < x < 1$, $x_0 = 1/2$. Newton’s method.
