The proof of Raabe's Test for absolute convergence In Introduction to Real Analysis second edition by Bartle & Sherbert's, there is a proof of Raabe's Test for absolute convergence. The problem is that I don't understand why some part of the proof is necessary. I will show you first the proof as it is in the book, and then explain what I don't understand. Part (a) of the test is as follows:
Raabe's Test: Let $X:=(x_n)$ be a sequence of nonzero real numbers. If there exists numbers $a>1$ and $K\in\mathbb{N}$ such that
$$\left|\frac{x_{n+1}}{x_n}\right|\leq 1-\frac{a}{n}\quad\text{for}\quad n\geq K,$$
then $\sum x_n$ is absolutely convergent.
Proof: If the inequality holds, then we have
$$k|x_{k+1}|\leq(k-1)|x_k|-(a-1)|x_k|\quad\text{for}\quad k\geq K$$
On reorganizing the inequality, we have
$$(k-1)|x_k|-k|x_{k+1}|\geq(a-1)|x_k|>0\quad\text{for}\quad k\geq K$$
from which we deduce that the sequence $(k|x_{k+1}|)$ is decreasing for $k\geq K$. If we add the last inequality for $k=K,\ldots,n$ and note that the left side telescopes, we get
$$(K-1)|x_K|-n|x_{n+1}|\geq(a-1)(|x_K|+\cdots+|x_n|).$$
This shows (why?) that the partial sums of $\sum|x_n|$ are bounded and establishes the absolute convergence of the series. Q.E.D.
Now, I don't see why it is important to show that the sequence $(k|x_{k+1}|)$ is decreasing. From the inequality
$$(K-1)|x_K|-n|x_{n+1}|\geq(a-1)(|x_K|+\cdots+|x_n|).$$
we have
$$(a-1)(|x_K|+\cdots+|x_n|)\leq (K-1)|x_K|\quad\text{for}\quad k\geq K$$
independently if $(k|x_{k+1}|)$ is decreasing or not since $n|x_{n+1}|>0$. So the partial sums of $\sum|x_n|$ are bounded anyway. Can someone explain me what I am missing?
 A: An alternate proof exists here 
http://my.safaribooksonline.com/book/engineering/9789332503632/1-sequences-and-series/head1_12_xhtml
A: Suppose $(k|x_{k+1}|)$ is not decreasing, which means there is a possibility that the terms in the sequence is getting larger and larger. In this case, you won't have a upper bound of the partial sum because $ (K-1)|x_K|-n|x_{n+1}|$ may be a negative number. 
A: Well, just because the partial sums are bounded doesn't mean a limit exits. However since $(k|x_{k+1}|)$ is decreasing, we can conclude that the partial sums are also decreasing and there is a theorem that states that any series that is bounded from bellow and decreasing is convergent. Hope this helps :)
A: In this proof the author desires to prove that the sequence of partial sums converges. He is trying to use the fact that a monotonic sequence in bound has a finite limit. Here the sequence of partial sums is decreasing monotonically and is bounded as shown in the inequality above. The sequence is also bounded by zero below since all terms are non-negative. Thus a finite limit exists and the series absolutely converges.
