Is the following criterion sufficient for Convergence?

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In summary: Since all of the terms are positive, the sequence of partial sums, 1, 1+ 1/2= 3/2, 1+ 1/2+ 1/3= 11/6, ... is an increasing sequence. That sequence is also bounded above (by 2) and so must have a limit. That limit is called the "Harmonic series" and is known to be infinite.In summary, we are discussing whether a bounded sequence, with the distance between successive terms decreasing, must necessarily converge. While intuitively this may seem true, it is in fact false as demonstrated by the example of the harmonic series. This sequence oscillates between two points and does not converge,
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Quantumpencil
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Homework Statement



Say {p_n} is a sequence in R, abs(p_n-p_n+1) -> 0, and {p_n} is bounded. Is it true that {p_n} must converge?



Homework Equations





The Attempt at a Solution



Intuition: Yes; in my attempts to find a counterexample I found sequences which diverged even though successive terms were closer (Partial sums of the Harmonic series), and were not bounded. From diagramming it seems to me that if you have the range of p_n bounded above and below, and the distance between successive terms must decrease, then the boundary will continue to close in from both sides until the sequence converges.

The only thing is I'm having trouble proving this without the assumption that the sequence is monotonic (I can't get a contradiction out of the existence of sup p_n or inf p_n)

So I'd like to make sure, before I continue with the proving, that this is actually true.

EDIT: I actually think this is false. Don't have a counter-example yet, but I think you might could have like, a sequence defined which oscillates around two points... I'm just not sure if I can do that without the oscillation being a convergent one.
 
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  • #2
You are wise to look hard for counterexamples before you start trying to prove. What about things like p_n=sin(sqrt(n))?
 
  • #3
Ah, so you do in fact need the Monotonicity assumption. That function is bounded. And as you increase the argument, the difference between successive terms decreases, but it still eventually oscillates back, bounded by 1 and -1 (sqrt 10000) and sqrt (10001) are closer together than sqrt (10) and sqrt (11), so even though the arguments start varying more slower, there is never any convergence.

You can show the divergence just by taking a divergent sub-sequence, and show the the closeness of successive terms with a bit of algebra.

Good stuff.

That explains my difficulty with trying to prove this... lol.
 
  • #4
Exactly. Oscillate between two points. Just do it slower and slower.
 
  • #5
The simplest and most famous example is the harmonic series:
[tex]\sum_{n=1}^\infty \frac{1}{n}[/tex]

The difference between successive terms is
[tex]\frac{1}{n}- \frac{1}{n+1}= \frac{1}{n(n+1}[/tex]
which goes to 0 (quadratically) as n goes to infinity.
 

FAQ: Is the following criterion sufficient for Convergence?

What is convergence?

Convergence is the process in which a series or sequence of numbers approaches a specific value as more terms are added. It is a concept in mathematics and is often used in calculus and analysis.

What is a criterion for convergence?

A criterion for convergence is a rule or condition that must be met in order for a series or sequence to converge. It is used to determine whether a series or sequence will approach a specific value or diverge.

What is the sufficient criterion for convergence?

The sufficient criterion for convergence is a rule or condition that, if met, guarantees that a series or sequence will converge. There are various criteria for convergence, such as the limit comparison test, ratio test, and root test.

Is the following criterion sufficient for convergence: "If the absolute value of each term in a series is less than or equal to the corresponding term in a convergent series, then the series is convergent"?

Yes, this criterion is known as the comparison test and is sufficient for convergence. It states that if the terms in a series are smaller than the terms in a known convergent series, then the series in question will also converge.

What should be done if the sufficient criterion for convergence is not met?

If the sufficient criterion for convergence is not met, then further analysis of the series or sequence is needed to determine if it converges or diverges. Other convergence tests, such as the integral test or alternating series test, can be used to determine the convergence or divergence of a series.

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