How to test if a sequence converges?

In summary, convergence in a sequence refers to the idea that as the terms in a sequence get closer and closer to a certain value, the sequence as a whole approaches that value as its limit. To determine if a sequence converges or diverges, you can use several methods such as the limit comparison test, the ratio test, or the root test. Absolute convergence refers to a series that converges regardless of the order in which its terms are summed, while conditional convergence refers to a series that only converges when its terms are summed in a specific order. No, a sequence can only converge to one limit and if it has multiple limits, it is said to be divergent. The epsilon-delta definition is a mathematical approach to testing convergence
  • #1
japplepie
93
0
Given a sequence, how to check if it converges?
Assume the sequence is monotonic but the formula that created the sequence is unknown.

My first thought was if:

seq(n+2) - seq(n+1) < seq(n+1) - seq(n) , is always true as n->infinity then it is convergent.

Or in other words, if the difference between two consecutive terms gets smaller and smaller, then it converges.

But it's not always true, an example of this is when seq(n) = log(n), where the difference between terms gets smaller and smaller, but it still diverges.

So my question is, is there a limit to how small the difference between consecutive has to be for it to converge?

Like, the difference between one term must be 1/2 the difference between the next term, etc
 
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  • #2
Seems difficult to me. As you know ##\Sigma x^n## converges for ##x<1## (if x >0). Your difference expression boils down to ##x(x-1)\lt (x-1) ## so 1/2 is far too strict...
 
  • #3
BvU said:
Seems difficult to me. As you know ##\Sigma x^n## converges for ##x<1## (if x >0). Your difference expression boils down to ##x(x-1)\lt (x-1) ## so 1/2 is far too strict...
I only used the 1/2 as an 'example answer'

We know that it's not enough for the difference of 2 consecutive terms getting smaller is not enough, so by how much smaller does it have to be for it to converge?
 
  • #4
To me it seems that a fixed ratio < 1 is enough. In other words a ratio ##1-\epsilon \ \ ## with ##\ \ \epsilon > 0##.

Your example has a ratio ## \displaystyle {n+1\over n+2} < 1## that isn't fixed. For all ##\epsilon > 0## an N can be found, such that ## \displaystyle {N+1\over N+2} > 1-\epsilon##.
 
  • #5
japplepie said:
Given a sequence, how to check if it converges?
Assume the sequence is monotonic but the formula that created the sequence is unknown

If the sequence is monotonic, then there is one necessary and sufficient condition for convergence: the sequence must be bounded.

Of course this can be difficult to prove, and writing it as a series ##x_n=\sum (x_{k+1}-x_k)## can help. Then you have various convergence criteria, typically amounting to showing that ##|x_{k+1}-x_k|\leq|y_{k+1}-y_k|## for some other sequence ##y_n## you already know converges.

The fact that ##x_{k+1}-x_k## is decreasing is not enough as your example shows, but if for instance ##\frac{|x_{k+1}-x_k|}{|x_k-x_{k-1}|}\leq a## where ##0<a<1##, then the sequence does converge (that condition is far from necessary though, if it doesn't hold you need to find something else).
 
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  • #6
Sure, but what the poster is after is a criterion expressed in the differences of subsequent terms

[edit] I realize that I have already bent that towards a ratio...
 
  • #8
Svein said:
A bounded sequence has one or more cluster points (https://en.wikipedia.org/wiki/Limit_point). This means that you can extract a convergent subsequence.
True but since we
Assume the sequence is monotonic
it is convergent in the present case.
 
  • #9
wabbit said:
If the sequence is monotonic, then there is one necessary and sufficient condition for convergence: the sequence must be bounded.

Of course this can be difficult to prove, and writing it as a series ##x_n=\sum (x_{k+1}-x_k)## can help. Then you have various convergence criteria, typically amounting to showing that ##|x_{k+1}-x_k|\leq|y_{k+1}-y_k|## for some other sequence ##y_n## you already know converges.

The fact that ##x_{k+1}-x_k## is decreasing is not enough as your example shows, but if for instance ##\frac{|x_{k+1}-x_k|}{|x_k-x_{k-1}|}\leq a## where ##0<a<1##, then the sequence does converge (that condition is far from necessary though, if it doesn't hold you need to find something else).
Can you please explain ##\frac{|x_{k+1}-x_k|}{|x_k-x_{k-1}|}\leq a## more, I couldn't 100% get it
 
  • #10
japplepie said:
Can you please explain ##\frac{|x_{k+1}-x_k|}{|x_k-x_{k-1}|}\leq a## more, I couldn't 100% get it
Sure - just staying in your case of an increasing sequence, under that assumption ##x_{k+1}-x_k\leq Ca^k## and ##x_n=x_0+\sum (x_{k+1}-x_k )\leq x_0+C\sum a^k## is bounded by the (finite since a<1) sum of a geometric series - so it must converge.
 

1. What is the definition of convergence in a sequence?

Convergence in a sequence refers to the idea that as the terms in a sequence get closer and closer to a certain value, the sequence as a whole approaches that value as its limit. In other words, the sequence “converges” to a specific value.

2. How can I determine if a sequence converges or diverges?

To determine if a sequence converges or diverges, you can use several methods such as the limit comparison test, the ratio test, or the root test. These methods involve evaluating the limit of a sequence and comparing it to known convergent or divergent series.

3. What is the difference between absolute convergence and conditional convergence?

Absolute convergence refers to a series that converges regardless of the order in which its terms are summed, while conditional convergence refers to a series that only converges when its terms are summed in a specific order. In other words, absolute convergence guarantees convergence, while conditional convergence does not.

4. Can a sequence converge to more than one limit?

No, a sequence can only converge to one limit. If a sequence has multiple limits, it is said to be divergent and does not converge to any specific value.

5. What is the role of the epsilon-delta definition in testing convergence of a sequence?

The epsilon-delta definition is a mathematical approach to testing convergence that involves finding a specific value of epsilon (a small positive number) for a given delta (a small positive number) such that all terms in the sequence are within a certain distance from the limit. This definition helps to rigorously prove the convergence of a sequence.

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