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By the agnecy of p 44 Definition 2.3.1, we find M > 0 such that $|a_n| \le M$ for all $n \in N$.

Not duplicate. Proof. By definition, given any $e >0$ there is an integer $N$ such that $|a_{n}- a_{m}| < e$ for all $n,\ m \geq N$. So $ |a_{n}| -|a_{m}| \leq |a_{n}-a_{m} | < e$ for all $n,\ m\geq N$.

1. The line overhead faults? It doesn't have my mod signs in red? Reverse Triangle Inequality invokes $ \color{Red}{|} \; |a_{n}| -|a_{m}| \; \color{Red}{|} \leq |a_{n}-a_{m} | < e$ ?

Take $m=N$ and transpose: $|a_{n}| < |a_{N}| +\epsilon$ for all $ n\geq N$.

2. Why take $m = N$? How to presage this step?

Thus, for all $n, |a_{n}|\ \leq\max\ \{|a_{1}|,\ .\ .\ .\ ,\ |a_{N-1}|,\ |a_{N}|\ +\epsilon\ \}. $

3. I don't understand 'thus.' We only proved $|a_{n}| < |a_{N}| +\epsilon$.
We never proved $|a_n| \le |a_i|$ for all $1 \le i \le N - 1$?

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    $\begingroup$ A user reading this question without reading your previous questions will not know what book you are referring to. Please include the name and author of the book in the text of the question. The title of the question is not part of the question itself, so you cannot assume that someone reading the question has read the title. Also, when you quote directly from a book, please format the question so that the direct quotation is clearly indicated. $\endgroup$ – Carl Mummert May 3 '14 at 20:02
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Before we get to details, do you understand the key idea? A sequence is Cauchy if all of the terms eventually cluster very close to each other. To show it's bounded, then, you just have to take one of those later terms and use it to take a bound just wide enough that captures all of the later terms. Now all you have to do is control the first (finitely many) terms of the series, which is easy because there are only finitely many of them.

  1. To go from the reverse triangle inequality to what is printed, transpose the difference (if necessary) so that the quantity inside the absolute values is positive.

  2. The reverse triangle inequality is true for all $m\geq N$, so we may take $m=N$. The reason we do this is to pick a single "reference point" for the tail of the sequence to control the behavior of all terms beyond $N$.

  3. You misunderstand the statement: The proof does not conclude that for all $n$, $|a_n|\leq |a_i|$ for ALL $1\leq i\leq N-1$. It concludes that for all $n$, $|a_n|$ is smaller than the largest of the first $N$ terms.

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$1. \; $ Yes. The line overhead does not contain your red signs. And the formal reverse triangle inequality states that $ | \ |a_{n}| -|a_{m}| \ | \leq |a_{n}-a_{m} | $. But is it not true that $ x \le |x| $ for any real number $x$? And hence should not $ |a_n | - |a_m| \le | \ |a_{n}| -|a_{m}| \ | $ and hence $ |a_n | - |a_m| \leq |a_{n}-a_{m} | $ ??

$2.\; $ What we need to do is to use the Cauchy criterion to find a bound for $a_n$ when $n$ is large enough. You do not need to pick $ m = N $. In fact you can pick any natural number not less than $N$. This is what we have obtained from our first argument:- Given a fixed positive quantity $e$ we have that for each $m, n \ge N$, $\;\; |a_n| \le 1 + |a_m| $. So we have a bound for $|a_n|$ as long as $m \ge N$ - this is the key point. (Note that our entire effort here is to find a bound for $|a_n|$). So $N$ is the obvious choice in place of the "variable", $m$. You can choose $2N, 3N, N + 1$ or $N^{1000}$ in place of $N$. It won't matter as long as the number you choose in place of $m$ is not less than $N$ for the inequality to work. How is this step presaged? Well it is not. All we've done is to fix $m$.

$3.\; $ So take a step back and look at what we have proven. As soon as we fix $m = N,$ $\;\; |a_N|$ becomes a constant non-negative number. And hence $ |a_N |+ e $ becomes a constant positive number. So we have proven that $ m, n \ge N \implies |a_n| \le e + |a_N| $. Since $m = N$ the condition $m \ge N$ is redundant and we have the very useful $$ n \ge N \implies |a_n| \le e + |a_N|$$

So we have bounded every element in the sequence beyond the $N$th index. Since the one's not taken care of -- $ |a_1|, |a_2|, ... |a_{N - 1}| $ -- are but a finite collection the value $M = \text{Max} \{ |a_1|, |a_2|, ... |a_{N - 1}| \}$ can be calculated. So we have that all $a_n$'s such that $n \lt N$ conform to $ |a_n| \le M $ as per the definition of $M$ and all $a_n$'s such that $n \ge N$ conform to $ |a_n| \le e + |a_N| $. Now let $M' = \text{Max} \{ M, (e + |a_N|) \}$. It should be clear that $|a_n| \le M'$ for each $n \in \Bbb N$ and hence $\{a_n\}$ is bounded.

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