Real Analysis | Cauchy Criterion for Series

We prove a few preliminary results for convergent series including some algebraic properties and the Cauchy criterion.
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[@nicholasroberts2933]

Michael, at 12:50, did you mean to say “contrapositive” instead of “converse”?

[@minime1235able]

Life would be so easy if the converse were true....

[@MichaelPennMath]

Yes, I should have said "contrapositive". Thanks for catching this!

[@BinodTharu-co4by]

Binod

[@hybmnzz2658]

There is a problem with the theorem at 10:20. It is not that the letters don't mirror each other; this is genuine.
The way the theorem is stated m+1 is the starting point of the finite sum while n controls the size of the finite sum. But since m,n >= N this would mean the size of the finite sum would have to be atleast N. This is more restrictive than the result we proved.
In the result we proved, m+1 is still an indicator of the starting point of the finite sum. But n-m represents the size of the finite sum. Since n,m >= N , This means the size of the finite sum can be an arbitrary natural number. This is similar but stronger than the theorem written on the board.

[@adrianamor8472]

The theorem statement is just wrong. If you are considering both m,n >= N then you should have |a_n+1 + ... + a_m| < eps.
The alternative version is to consider n >= N and p >= 1, then you have |a_n+1 + ... + a_n+p| < eps.
This statement is mixing the two above, which is just not Cauchy criterion (perhaps still holds but i'd consider it wrong).
Also, Just take p = 1 for the corollary proof, taking n = m + 1 doesn't give you a_n, it gives you a_m+1 + ... + a_(2m+1).
Taking p=1 you get |a_n+1| < eps which is not |a_n| but you could just pick K = N+1 and then for all n >= K => |a_n| < eps

[@MrCentrax]

Found your chanell when I was studying some basic number theory and I have to say your's one of the best math channels on CZcams. Great work

[@EmilianoPM6754]

There is an error in the theorem, as n does not necessarily have to be greater than N given the form of the theorem that is written. It would be fine if the sum inside the absolute values ended in a_n and not a_(n+m). In the form given we should only ask for n>=1.

[@thenewest1]

Hey Michael, big fan of the RA series, was wondering how far you were planning on going with this series e.g. just a few videos or maybe following a textbook. I'm Preparing for the math GRE and brushing up on RA with your videos is a godsend.

[@MichaelPennMath]

I am making these videos to support a full semester course in Real Analysis that I am teaching this Fall.

[@BinodTharu-co4by]

Binod

[@mohameddaoud4885]

14:15 why would it collapse to |a_n|, I believe it should be |a_n, a_n+1, a_n+2, ...... a_2n-1|? Hope to clear my confusion. Thank you for everything

[@stefanschroder4694]

You are absolutely right :)

[@davidmoss9926]

Referencing Abbott I believe the finite sum should be a_m+1 +...+a_n rather than a_m+1 +...+ a_m+n

[@thesecondderivative8967]

​@@davidmoss9926 I believe so as well.

[@thesecondderivative8967]

14:13 I think the sequence of terms in the absolute value should end in n and not m+n. That way, it would collapse.

[@jimallysonnevado3973]

Why if we set n=m+1, the sum collapses shouldn’t it be |(a_m+1 +a_m+2+.....a_2m+1)|?

[@jimallysonnevado3973]

There is a mistake in the index of the stated theorem, it should be |a_m+1+.....+a_n| not |a_m+1....+a_m+n|.

[@asht750]

Can you also make more videos on advanced topics in Real Analysis such as the Contraction Principle, and so forth?

[@goodplacetostop2973]

14:36

[@victorserras]

Incredible stuff! Thanks.

[@youssefbouhtouch6050]

I have a question :
if we have two polynomials p and q in Q[x]
and there exists a complex number z such that :
p(z)=q(z)=0 (z is a root fo each of p and q)
can we say that gcd(p,q) in Q[x] is not a constant and please why

[@thenewdimension9832]

Best explanation!

[@zakthayer9315]

12:20, appreciate the effort 😂

[@JB-ym4up]

You say its koshi but can you parve it?

[@euclidtheorem4817]

XD I'm wheezing.

[@BinodTharu-co4by]

Binod

[@________6295]

14:37