How to test if a sequence converges?

  • Thread starter japplepie
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  • #1
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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
BvU
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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
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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
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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
wabbit
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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
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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
wabbit
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  • #9
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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
wabbit
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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.
 

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