Convergence and divergence of series and sequences

In summary, convergence refers to the behavior of a sequence or series approaching a specific value as the number of terms increases, while divergence indicates that it does not settle at any particular value. A convergent series has a finite sum, whereas a divergent series can either grow indefinitely or oscillate without approaching a limit. Key tests for convergence include the ratio test, root test, and comparison test, while divergence can often be identified through these tests or by recognizing patterns in the terms. Understanding these concepts is crucial in mathematical analysis and applications in various fields.
  • #1
chwala
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TL;DR Summary
I am going through the notes- refreshing. Would like to share my insight and probably clear some doubts highlighted in purple below.
Theorem
1. If a series ##{a_n}## converges, then the sequence ##{a_n}## converges to ##0##.
Now, the contra does not apply, and my question is why? i.e if the the sequence ##{a_n}## converges to ##0## then the series may or may not converge correct? and if it does not converge to ##0## then it diverges i.e for e.g ## \sum_{n=1}^\infty \dfrac{1}{2n}##

##\lim_{n→∞} \dfrac{1}{2n}=0## but the series diverges.

Secondly, in establishing convergence of series- we can look at a series on the context of a function i.e for e.g we can look at

## {a_n}=\sum_{n=1}^\infty \dfrac{1}{2n}## as ##f(x) = \sum_{x=1}^\infty \dfrac{1}{2x}##.

Thirdly, it seems to me that the the integral test is the more general approach to use in establishing on whether or not a series converges or diverges? other than the other available tests i.e ratio, comparison, p-test. Correct?
 
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  • #2
Consider the sequence 1/1, 1/2, 1/3... it converges to zero

but the series 1/1+1/2+1/3+1/4... does not

1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+...

since it is bigger than this series where we replace some terms with power of two terms that are smaller ie 1/3 -> 1/4 and 1/5, 1/6, 1/7 are replaced by 1/8... then group terms to get a series of 1+1+1...

1/1+(1/2+1/4+1/4)+(1/8+1/8+1/8+1/8+1/16+1/16+...) + ...
1+. (1) + (1) + ...

more on the harmonic series is here:

https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)
 
  • #3
jedishrfu said:
Consider the sequence 1/1, 1/2, 1/3... it converges to zero

but the series 1/1+1/2+1/3+1/4... does not

1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+...

since it is bigger than this series where we replace some terms with power of two terms that are smaller ie 1/3 -> 1/4 and 1/5, 1/6, 1/7 are replaced by 1/8... then group terms to get a series of 1+1+1...

1/1+(1/2+1/4+1/4)+(1/8+1/8+1/8+1/8+1/16+1/16+...) + ...
1+. (1) + (1) + ...

more on the harmonic series is here:

https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)
I get you @jedishrfu ...i understand that. My question probably should be why is there not a theorem on that effect.
 
  • #4
chwala said:
I get you @jedishrfu ...i understand that. My question probably should be why is there not a theorem on that effect.

A counterexample which shows that a proposition is not true is a proof of the theorem that the proposition is false.
 
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  • #5
There are times in mathematics when we can't explain why, but know there is a counter-example that suffices to shut down the proposition.

There are also cases in systems of mathematics where something is undecidable, as we can't prove or disprove it. (Godel's theorem)
 
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  • #6
chwala said:
TL;DR Summary: I am going through the notes- refreshing. Would like to share my insight and probably clear some doubts highlighted in purple below.

Theorem
1. If a series ##{a_n}## converges, then the sequence ##{a_n}## converges to ##0##.
Now, the contra does not apply, and my question is why?
See the counterexample of the harmonic series as mentioned by @jedishrfu. Why? Some sequences aren't simply fast enough on their way to zero. Even steps of ##1/n## add up to infinity.

chwala said:
i.e if the the sequence ##{a_n}## converges to ##0## then the series may or may not converge correct?
Yes. See above.
chwala said:
and if it does not converge to ##0## then it diverges i.e for e.g ## \sum_{n=1}^\infty \dfrac{1}{2n}##

##\lim_{n→∞} \dfrac{1}{2n}=0## but the series diverges.
Yes.
chwala said:
Secondly, in establishing convergence of series- we can look at a series on the context of a function i.e for e.g we can look at

## {a_n}=\sum_{n=1}^\infty \dfrac{1}{2n}## as ##f(x) = \sum_{x=1}^\infty \dfrac{1}{2x}##.

That makes no sense. Your function is the same as the series. What did you gain by renaming the summation index except confusion? You could consider the integrals, e.g.
$$
\int_{1}^{N+1}\dfrac{1}{x}\,dx <\sum_{k=1}^{N}\dfrac{1}{k}<1+\int_{1}^{N}\dfrac{1}{x}\,dx
$$
but you must be careful. A summation and an integral are different concepts and you cannot simply replace the symbols.

chwala said:
Thirdly, it seems to me that the the integral test is the more general approach to use in establishing on whether or not a series converges or diverges? other than the other available tests i.e ratio, comparison, p-test. Correct?
See here: https://www.physicsforums.com/insights/series-in-mathematics-from-zeno-to-quantum-theory/
 
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  • #7
chwala said:
Thirdly, it seems to me that the the integral test is the more general approach to use in establishing on whether or not a series converges or diverges? other than the other available tests i.e ratio, comparison, p-test. Correct?
The integral test is applicable in this case and nothing more. The most perfect or the most general test for convergence or divergence of a series does not exist. (https://en.wikipedia.org/wiki/Convergence_tests)
 
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  • #8
Gavran said:
The integral test is applicable in this case and nothing more. The most perfect or the most general test for convergence or divergence of a series does not exist. (https://en.wikipedia.org/wiki/Convergence_tests)
If you analyze the proofs of those tests, you will find out that they are all based on the comparison test, sandwiching.
 
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