Lp is complete proof understanding Hi I am having some trouble understanding the proof which is written in Schillings' measure theory
Statement: The spaces $L^{p} (\mu)$ are complete for $p \in [1, \infty]$. 
The proof goes like this
Assume $p \in [1, \infty)$. The main difficulty is to identify the limit $u$
By the definition of cauchy sequence we find numbers
$ 1 < n(1) < n(2) < ... < n(k)$
such that
$||u_{n(k+1)} - u_{n(k)}||_{p} < 2^{-k}$
To find $u$ we turn the sequence into a series by 
$u_{n(k+1)} = \sum^{k}_{i = 0} (u_{n}(i+1) - u_{n}(i)) , \ u({n_{0}}) = 0$
And as the limit $k \rightarrow \infty $ would formally be
$u :=\sum^{\infty}_{i = 0} (u_{n}(i+1) - u_{n}(i)) $
Now if we can make sense of this infinite sum. Since
$||\sum^{\infty}_{i=0} |u_{n(i+1)} - u_{n(i)}| ||_{p} \leq  \sum^{\infty}_{i=0} ||u_{n(i+1)} - u_{n(i)}||_{p} \leq ||u_{n}(1)||_{p} + \sum^{\infty}_{i=1} \frac{1}{2^{i}} (*) $
I don't see where this last inequality $(*)$ is comming from and why does that imply $u$ is absolutely convergent?
Edit:
Let us show that $u = L^p - lim_{k \rightarrow \infty} u_{n(k)}$
For this observe that by the ordinary triangle inequality and from above that 
$||u - u_{n(k)}||_{p} =  || \sum^{\infty}_{i = k} (u_{n (i+1)} - u_{n(i)})||_{p}$, I don't really see where is this equality comming from?
and after some applications of the triangle inequality he gets that 
$\sum^{\infty}_{i = k} || u_{n(i+1)} - u_{n(i)}||_{p} \rightarrow 0 \ as \ k \rightarrow \infty$ I am wondering how come this goes to 0 as k goes to infinity?
 A: The first $\leq$ comes from moving the norm inside the sum. The second is this:
\begin{align*}\sum\limits_{i=0}^\infty\|u_{n(i+1)}-u_{n(i)}\|_p&=|u_{n(1)}-u_{n(0)}\|_p+\sum\limits_{i=1}^\infty\|u_{n(i+1)}-u_{n(i)}\|_p\\
&=\|u_{n(1)}\|_p+\sum\limits_{i=1}^\infty\|u_{n(i+1)}-u_{n(i)}\|_p\\
&\leq \|u_{n(1)}\|_p+\sum\limits_{i=1}^\infty2^{-i},
\end{align*} using that $u_{n(0)}=0$ and the choice of the subsequence.
EDIT: Let me add in a full proof. Given $(u_n)$ Cauchy in $L^p$, we can pass to a subsequence and assume that $$\|u_{n+1}-u_n\|\leq 2^{-n}.$$ Consider the sum $$u_1(x)+\sum\limits_{n=1}^\infty (u_{n+1}(x)-u_n(x)).$$ We can dominated the partial sums of this series by $g_m(x)=\sum\limits_{n=1}^m|u_{n+1}(x)-u_n(x)|.$ Note that $(g_m)$ satisfies that $0\leq g_1\leq g_2\leq\cdots,$ and $\|g_m\|_{L^p}\leq \sum 2^{-n}\leq 1.$ By the monotone convergence theorem, $g_m\nearrow g$ a.e. and in $L^p$ norm. This implies that $$u_1(x)+\sum\limits_{n=1}^\infty (u_{n+1}(x)-u_n(x))$$ converges a.e. to a limit, call it $u$, and applying the dominated convergence theorem yields that $u_n\rightarrow u$ in $L^p$ norm. At the end, we are using that $u_k=u_1+\sum\limits_{n=1}^{k-1}(u_{n+1}-u_n).$ 
For your new question, that's just pretty much what it means for that series to converge (which we know it does).  
