Why can we consider this subsequence in $L^p$? I am trying to understand a step in the following proof of completeness of $L^p$ in Stein-Shakarchi's Functional Analysis. (See the proof on page 5 of the link or at the end of this post.)
At the beginning of the proof, it is said that

Let $\{f_n\}_{n=1}^\infty$ be a Cauchy sequence in $L^p$, and consider a subsequence $\{f_{n_k}\}_{k=1}^\infty$ of $\{f_n\}_{n=1}^\infty$ with the following property $\|f_{n_{k+1}}-f_{n_k}\|\le 2^{-k}$ for all $k\geq 1$.

Question: Why can the sequence be considered as it is?
On a YouTube video, it explains about a similar subsequence.
I still don't understand why for  $n,m>n_k$ $\Vert f_{n}-f_{m}\Vert_p\implies \Vert f_{n_k}-f_{n_{k+1}}\Vert_p$, thus an increasing subsequence. Why is it justified to make $n$ to depend on $k$, $n_k$?




 A: If $\{f_n\}_n$ is Cauchy, for each $k\in\Bbb N$ we can find an $N_k$ (depending on the $k$ we just chose) such that
$$ \|f_n - f_m\| < 2^{-k} \quad\text{ whenever } n,m\geq N_k.$$
For $k+1$, we likewise get
$$ \|f_n - f_m\| < 2^{-k-1} \quad\text{ whenever } n,m\geq N_{k+1}.$$
Now, both inequalities will hold for $n,m \geq \max\{N_k,N_{k+1}\}$, so we can choose $N_{k+1} > N_k$, and similarly we can choose an $N_{k+2} > N_{k+1} > N_{k}$ for $2^{-k-2}$ so that
$$\|f_{N_k} - f_{N_{k+1}}\| < 2^{-k}\quad\text{and}\quad \|f_{N_{k+1}} - f_{N_{k+2}}\| < 2^{-k-1}$$
by choosing $n = N_k$ and $m = N_{k+1}$, etc in the first inequalities above.
A: The authors mention at the beginning of the proof that 

The argument is essentially the same as for $L^1$ (or $L^2$); see Section 2, Chapter 2 and Section 1, Chapter 4 in Book III. 

It is said clearly there (see also a snapshot at the end) that

The existence of such subsequence is guaranteed by the fact that $\|f_{n}-f_{m}\|\leq \epsilon$ whenever $n,m\geq N(\epsilon)$, so that it suffices to take $n_k=N(2^{-k})$.

[Added for elaboration.]  
There exists an integer $N(2^{-1})>0$ such that for all $n,m\geq N(2^{-1})$,
$$
\|f_{n}-f_{m}\|\leq 2^{-1}\tag{1}.
$$
There exists an integer $N(2^{-2})>N(2^{-1})$ such that for all $n,m\geq N(2^{-2})$,
$$
\|f_{n}-f_{m}\|\leq 2^{-2}\tag{2}.
$$
There exists an integer $N(2^{-3})>N(2^{-2})$ such that for all $n,m\geq N(2^{-3})$,
$$
\|f_{n}-f_{m}\|\leq 2^{-3}\tag{2}.
$$
... so on and so forth. 
Now, let $n_1=N(2^{-1})$, $n_2=N(2^{-2})$, $n_3=N(2^{-3})$, $\cdots$. 
Since $n_1,n_2\geq N(2^{-1})$, we have by (1)
$$
\|f_{n_2}-f_{n_1}\|\leq 2^{-1}.
$$
Since $n_2,n_3\geq N(2^{-2})$, we have by (2)
$$
\|f_{n_3}-f_{n_2}\|\leq 2^{-2}.
$$
Since $n_3,n_4\geq N(2^{-3})$, we have by (3)
$$
\|f_{n_4}-f_{n_3}\|\leq 2^{-3}.
$$
... so on and so forth. 

The following is a snapshot of the beginning of the proof for completeness of $L^1$ in Stein-Shakarchi's Book III (page 70 Theorem 2.2).



