
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
November 11th, 2017, 07:47 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 07:50 AM. 
November 11th, 2017, 09:41 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,174 Thanks: 1143 
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. 
November 11th, 2017, 11:00 AM  #3 
Newbie Joined: Oct 2017 From: Here Posts: 19 Thanks: 0 
Thank you

November 11th, 2017, 05:24 PM  #4  
Member Joined: Jan 2016 From: Athens, OH Posts: 91 Thanks: 47  Quote:
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, 01:12 PM  #5  
Newbie Joined: Oct 2017 From: Here Posts: 19 Thanks: 0  Quote:
 
November 19th, 2017, 01:29 PM  #6  
Senior Member Joined: Aug 2012 Posts: 2,082 Thanks: 595  Quote:
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 01:33 PM.  

Tags 
definition, diverge, infinity, sequence, showing 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
proof that sequence > infinity  PrototypePHX  Real Analysis  3  July 9th, 2013 01:39 PM 
Showing convergence/divergence of sequence with Cauchy  MageKnight  Real Analysis  16  May 18th, 2013 02:33 PM 
Limit at Infinity of an Oscillating Sequence  Code2004  Calculus  2  January 18th, 2013 08:56 PM 
Some feedback on showing that a sequence is convergent?  tach  Real Analysis  6  December 14th, 2009 06:57 PM 
Does the Sequence Converge or Diverge?  veronicak5678  Calculus  2  November 4th, 2008 10:18 AM 