A nice "advanced" calculus result

[@thatdude_93]

For the "=>" direction you can also argue like this: sum a_n converges to some value a < infnity. Then e^a is also finite. But e^a=e^(sum a_n)=product e^a_n >=product(1+a_n), since it is well known that e^x>= 1+x for all real x. Hence the product must also converge, since it is bounded from above and clearly increasing.

[@pwmiles56]

I think you could extend to = x/2 for 0

[@thatdude_93]

@@pwmiles56 well observed

[@bjornfeuerbacher5514]

This result also gives a very nice and fast proof that the harmonic series diverges, since obviously 1/n > 0 and the infinite product (1 + 1/n) diverges. :)

[@TheEternalVortex42]

I don't think it's immediately obvious that this infinite product diverges?

[@عبداللهعبدو-ع1ج]

​@@TheEternalVortex42 the partial product in this case is equal to n +1 try to prove it by induction

[@juyifan7933]

@@TheEternalVortex42 It is, the product "telescopes", try it out.

[@bjornfeuerbacher5514]

@@TheEternalVortex42 Consider the first few terms (starting with n = 1, obviously). The product is 2 times 3/2 times 4/3 times 5/4 times 6/5 ...... Do you see now why it's obvious that this diverges? (It's essentially the same argument as for the product in the first example in the video, starting at about 11:15.)

[@bjornfeuerbacher5514]

@@angelmendez-rivera351 Point taken. :)

[@jakobr_]

For the reverse direction, could we try to “distribute” through the entire product? The result would be 1 plus the sum of all a_n plus the sum of every product of two different a_n plus the sum of every product of 3 different a_n etc., all of which are positive, so that product is greater than the sum of all a_n, therefore the sum of a_n converges.

[@ranitacab]

Nice !

[@MrDestroys]

That truly is a nice advanced calculus result

[@CM63_France]

Hi, 9:30 : and tends to 0 as n tends to infinity 16:54 : it was not a good place to stop the image because the sound was not finished 😄

[@somecreeep]

A question I've always been curious of regarding these two convergent series is "What properties about a sequence can be known if you only know what the sum and product converge to?"

[@ethanbartiromo2888]

For a really easy counter example of the second direction can’t we just do a_n = -1 for all n which gives us the product of infinite identically zero values, so the product converges to 0, but the sum would be the infinite sum of -1’s which would be -infty

[@memesThatDank]

keep up Michael!

[@mathpuzzles6352]

Thanks for making it!

[@arthurfibich112]

You can loosen the restriction of a_n>0 that caused everything to break in both failure cases, if you instead assume absolute convergence on the left side and unconditional convergence on the right side. Then you even get that this result holds for complex numbers a_n and for every possible branch of the logarithm that you use in between. Great job btw. on the video

[@RobsMiscellania]

A lovely result about certain sequences of real numbers. Well done video.

[@pierreabbat6157]

Both counterexamples have the terms approaching 0 like -1^n/√n. Can you make a condition like "a[n]>0 or |a[n]| shrinks faster than 1/√n or |a[n]| shrinks slower than 1/√n"?

[@discrete3511]

Excellent 👍🏻

[@ddognine]

Probably unrelated but this reminds me of a problem in my calculus text called the infinite paint can where the integral of 1/x from 1 to inf diverges along with the surface area formed from revolving 1/x around the x-axis. However, the volume from the revolution converges which means a paint can shaped as such cannot hold enough paint to paint its exterior. Definitely a weird result to wrap your head around if you can! (pun intended)

[@Kapomafioso]

I love that, too! I think it's interesting to push this "paradox" a bit further and investigate if we can actually resolve it. I prefer this "solution": ask, what it means to "paint" something? To me it means to apply a layer of a certain thickness d on that object, so that (approximately) volume = surface area*d is used. However, how does this work out with the shape you're describing? Obviously, at some point, no matter how small you pick d to be, the shape is going to narrow down even more past a certain point, so we can either conclude, that it is considered painted when the shape is filled with paint, or, we can conclude that the point wouldn't even be able to flow all the way to the back of the shape (even given infinitely long time), because the paint has a finite viscosity that wouldn't allow it to thin any further.

[@IsomerSoma]

I am not exactly sure as i have only just started studying measure and integration theory, but the reason for this might be the difference between a n-dim volume in n-dim space and a lower dimensional surface in a higher dimensional space being measured differently. It only may seem paradoxical if you dont know the mathematical machinery behind it and try to apply abstract mathematical results to the world directly. Imo there's nothing to wrap your head around but that mathematics shouldn't be mystified to contain truths about the natural world before using it as a tool to model nature appropriately.

[@Geffde]

I feel like it’s kind of cheating to jump directly to complex numbers when the previous restriction was just positive reals. Is there a counterexample if a_n contains negative reals?

[@pandabearguy1]

If you ain't cheatin you ain't tryin

[@ST-yb2el]

Example in Math Stack exchange question 1921045. a_2n = 1/sqrt(n). a_2n+1 = -1/1+sqrt(n). Product is 1 but series diverges.

[@chrissch.9254]

I wonder whether the first example can be considered a counter example at all as the theorem requires all coefficients a_n to be positive; therefore this is a theorem of real calculus as complex numbers can‘t be ordered…

[@alexanderbasler6259]

Just let a_n=-1. Then the product is 0, but the sum over the a_n trivially diverges.

[@Happy_Abe]

@@alexanderbasler6259 quite the simple counterexample!

[@carstenmeyer7786]

You can also prove both directions using the arithmetic/geometric/harmonic-mean-inequality (AM-GM-HM), with the product taking the place of GM in the middle. The proof is very symmetrical, and it is very nice to use the entire AM-GM-HM-inequality for once, not just AM-GM^^

[@AlexBesogonov]

There's a much easier way. Just take a logarithm of the product series, it'll transform it into a sum of logarithms. Then you just need to prove that Sum(ln(1+a_n)) converges if Sum(a_n) converges.

[@themathhatter5290]

Is there a sum the converges but relevant product does not, without going into the complex plane?

[@darkmask4767]

16:53 good place to stop

[@admink8662]

Wow, interesting result.

[@r.maelstrom4810]

Why expand the product to 1+ i/sqrt(2N) instead of 1+ i/sqrt(N)? The result after telescopic cancellation is N+1 which is infinity anyways...

[@driksarkar6675]

Does the convergence of the product of a_n necessarily imply the convergence of the sum of ln(a_n)? What if the product is 0?

[@ojas3464]

👍The sum on the left at 14:30 starts with oneth term, so zeroth term does not exist

[@Alex_Deam]

He's expanding the e as a Taylor series, he's not expanding sum of all the exponential terms of the sequence. So there is a 0th term to the Taylor series.

[@ojas3464]

@@Alex_Deam Thanks for pointing the nuance, each term needs to be first evaluated and then only the big Sigma, since operator precedence is higher for exponentiation than for multiplication, then for addition☺

[@ismaelperez9020]

Wait a minute. I can swap out operators (sums and products) and still use these theorems? Can I swap them out for other operators? What theorem says that I can do this or not?

[@와우-m1y]

asnwer=1 isit pier isit

[@aweebthatlovesmath4220]

6:00 it might not if product is zero

[@dlevi67]

How would the product of a(n = 0 to ∞) be zero when a(n) > 0?

[@seneca983]

That would be a divergence, wouldn't it?

[@radadadadee]

4:12 this is too messy for me, I can't keep watching. You cannot be rigorous or liberal whenever it pleases you.

[@seneca983]

Does the result apply in the forward direction if we assume that a_n is real but not necessarily positive?

[@arthurfibich112]

You have to assume absolute convergence, but then yes.

[@YoutubeBS]

@@arthurfibich112 and if not? Is there a counter example?

[@jkid1134]

This one smells like a lemma

[@radadadadee]

your mom smells like a lemma

[@uzytkownik9616]

Can anyone tell me where i can find math problems like this?😂

[@radadadadee]

in a math book

[@molteriet]

Excellent clickbait, leaving the a_n > 0 out of the thumbnail 😅

[@tomholroyd7519]

Can somebody point me to a proof of L'Hôpital's rule using smooth infinitesimal analysis (SIA) aka SDG I guess (dx is nilsquare). No limits 😜

[@s4623]

What about if aₙ ≥ 0 instead of just aₙ > 0? How would that be different?

[@RandomBurfness]

If a_n = 0 for some, or possibly even infinitely many n's, then you can just create a subsequence of the a_n's that aren't equal to 0 and you would get the same sum because you're just removing zeroes, and you'd get the same product because if you have a_n = 0 then 1 + a_n = 1 and multiplying by 1 changes nothing. Thus, the theorem is only stated for a_n's strictly larger than 0.

[@seneca983]

@@RandomBurfness I think the case of some a_n being 0 was left out because it would create complications in the limit comparison test if the denominator could be 0.