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 November 11th, 2017, 06:47 AM #1 Newbie   Joined: Oct 2017 From: Here Posts: 19 Thanks: 0 Showing a sequence does not diverge to infinity using definition Is there a definition for a sequence that does not diverge to infinity? Can I negate the definition of divergence to infinity of sequence, like I can do with convergence? What I mean is, convergent sequence definition is: "A sequence (an) is said to converge to L provided that for each ϵ>0 there exists a number N such that n>N implies |an−L|<ϵ" Now, saying that the limit is not L by definition is: "There exists an ϵ>0 such that for every number N, there exists n≥N with |an−L|≥ϵ" But what is the divergence to infinity equivalent? I know that the definition of divergence to infinity is: "A sequence (an) diverges to ∞ if for each real number M>0 there exists a N such that (an)>M for all n⩾M" What is the negating of the above statement? It does not seem to work the same as the convergence definition. What do I need to show if I want to prove that a limit of a sequence does not diverge to infinity using the definition? Thanks Last edited by Mathmatizer; November 11th, 2017 at 06:50 AM. November 11th, 2017, 08:41 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 A sequence doesn't diverge to infinity if it is bounded. If you can show there exists some real number $M\geq 0$ such that $|a_k| \leq M,~\forall k \in \mathbb{N}$, you will have shown that $a_n$ doesn't diverge to plus or minus infinity. You will NOT, however, have shown that $a_n$ converges. Thanks from topsquark, v8archie and Mathmatizer November 11th, 2017, 10:00 AM #3 Newbie   Joined: Oct 2017 From: Here Posts: 19 Thanks: 0 Thank you November 11th, 2017, 04:24 PM   #4
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Quote:
 if for each real number M>0 there exists a N such that (an)>M for all n⩾M
Here's your definition rewritten slightly (typo n>M corrected):
for all $M>0$ there exists a positive integer N such that for any $n\geq N$, $a_n>M$.

Now remember the negation of "for all" is "there exist" and the negation of "there exist" is "for all". So just negate each quantifier in the above and negate the statement:

There is $M>0$ such that for any positive integer $N$ there exists $n\geq N$ such that $a_n\leq M$.

Note this does not say that the sequence is bounded. Try and think of a sequence that satisfies the above negation, but is not bounded.

Finally, with a little practice, you can "mechanically" negate any quantified statement. November 19th, 2017, 12:12 PM   #5
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 Originally Posted by romsek A sequence doesn't diverge to infinity if it is bounded. If you can show there exists some real number $M\geq 0$ such that $|a_k| \leq M,~\forall k \in \mathbb{N}$, you will have shown that $a_n$ doesn't diverge to plus or minus infinity. You will NOT, however, have shown that $a_n$ converges.
But what if the sequence is (-2)^n? Then it is not bounded, and doesn't diverge to infinity, so showing a sequence is bounded isn't always the solution November 19th, 2017, 12:29 PM   #6
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 Originally Posted by Mathmatizer But what if the sequence is (-2)^n? Then it is not bounded, and doesn't diverge to infinity, so showing a sequence is bounded isn't always the solution
Your original question was: "Is there a definition for a sequence that does not diverge to infinity?"

The answer is that if the sequence is bounded, it does not diverge to infinity.

But if the sequence is not bounded, it might or might not diverge to infinity as your example shows.

The general logical form is that just because $P \implies Q$, we can not conclude that $\neg P \implies \neg Q$. That would be a fallacy.

So just because "bounded" implies "doesn't diverge to infinity," we can NOT conclude that unbounded implies diverges to infinity.

Last edited by Maschke; November 19th, 2017 at 12:33 PM. Tags definition, diverge, infinity, sequence, showing Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post PrototypePHX Real Analysis 3 July 9th, 2013 12:39 PM MageKnight Real Analysis 16 May 18th, 2013 01:33 PM Code2004 Calculus 2 January 18th, 2013 07:56 PM tach Real Analysis 6 December 14th, 2009 05:57 PM veronicak5678 Calculus 2 November 4th, 2008 09:18 AM

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