7

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?

Spenser
  • 19,469
  • 1
    It isn't important, you've not been missing anything. What you need is the inequality $\lvert x_K\rvert + \dotsc + \lvert x_n\rvert \leqslant C$ for some constant $C$ and all $n \geqslant K$. Here one proves that inequality for $C = \frac{K-1}{a-1}\lvert x_K\rvert$, and in that proof one coincidentally proves that the sequence $(k\lvert x_{k+1}\rvert)$ is (eventually) strictly decreasing. Such things are often interesting properties, maybe that's why the authors pointed it out explicitly, but it's not really important for the proof. Well, on another level it is important, since it's the … – Daniel Fischer Jun 23 '16 at 13:43
  • 1
    … inequality $(k-1)\lvert x_k\rvert - k\lvert x_{k+1}\rvert \geqslant (a-1)\lvert x_k\rvert$ that gives the upper bound $C$, and that inequality also shows that $(k\lvert x_{k+1}\rvert)$ is (eventually) decreasing. But there's no intrinsic need to mention the latter fact. – Daniel Fischer Jun 23 '16 at 13:43

4 Answers4

3

An alternate proof exists here http://my.safaribooksonline.com/book/engineering/9789332503632/1-sequences-and-series/head1_12_xhtml

MathMan
  • 8,974
  • 7
  • 70
  • 135
1

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 :)

Surfrider
  • 21
1

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.

Lizzy Wong
  • 34
  • Sorry. This may not be the exact reason. I think the author just want to emphasize $(K-1)|x_K|$ is the largest term so that you can use it to control the partial sum. – Lizzy Wong Aug 10 '13 at 18:57
  • This cannot happen. If $(K-1) |x_K| - n |x_{n+1}|$ is negative, then the whole inequality does not hold, since the right-hand side is non-negative. – Bodo Manthey Dec 14 '21 at 20:32
-1

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.

user194769
  • 1