12:12 AM
How do I get let back into MO earlier than 6 or so months from now when it auto-lets me in?
My questions look fine. I don't get it
They need to be less opaque about all of this, especially since we're content creators for there fucking site and we're not getting paid
[/rant over]. Solution; wait 6 months :0

I figured it out
I don't wait
@DanielDonnelly
I ask questions on mathoverflow whenever I want
it's because they know I'm research level
so they downvote but under their breath they whisper "yeah that's research level"
I got +5 on a post about tessellating $\Bbb R^3$ with orbifolds of a given type related to an Ising type model
lattice sites and all that jazz

12:43 AM
@JohnZimmerman that's neat. Well the system liked yo shit :)
And my shit is shit
@JohnZimmerman do you like combinatorics / graph problems?

I think the system just recognizes game

We should work together on something an open problem out there
I'm pretty well introduced to graph canonization rn. If you could efficiently canonicalize your space of graphs (the likely graphs in your application), then all you need to do to do a subgraph-isomorphism search is hold everything in your host language's built-in map or dictionary.
It's freakin cool. Near O(1) hashing to search for all graphs "in the database" isomorphic to yours
But the canonical form approximating would bring that up to $O(n^2)$ at least, where $n = |V|$ the number of vertices

@DanielDonnelly what open problem do you want?

A canonical form is some unique string or integer token you can pass around. As long as the recipient has your decoder function, then they know what graph you are taling about (up to automorphism of course)
I want to build a custom graph database for hosting commutative diagrams. But the graph theoretic stuff gets really deep into abstract algebra / group theory
My graphs can be up to say $256$ nodes and $16$ incident directed edges in/out of each node. But perhaps we should make these numbers smaller as the naive canon. algorithm on a $256$ node graph takes $\sim 256! \times C$ some constant
hours
days
months
the expiration of the galaxy even
So we need to use symmetry to cut down the problem

@DanielDonnelly sent to fruitful approach

12:49 AM
We no more than like a million steps in our high-level algorithm if you unroll all the loops
@JohnZimmerman got your email. Please read / scan the Wiki artical on graph canonization. Shoot me some questions about it and I can try to answer. I've been searching around
I no longer think $P = NP$. However we can "do our best" with what computer resources we have
It's as though when you're working on these problems, you reduce to another problem then another then another, and each new problem is just as hard to solve!!! They end up becoming some certain graph coloring problem that is still open, etc. It's a whackamole that one cannot beat seemingly

1:29 AM
@JohnZimmerman sorry my laptop lost power please check for new invite

1 hour later…
2:43 AM
@JohnZimmerman
:)
I have an idea for a website
Mostly human edited and not checked with a proof engine; Like a state machine of powerpoint slides. So you have a formula on one page. Then a collection of buttons. One of them says "Take the mellin transform" and that brings you to another human edited page, full with remarks
You can also move backward with the inverse transform, but there will be all sorts of buttons and you can sort through them by popularity, how many times it's been clicked by you, or by date created, etc. This is a really good design for an all-platform app that even goes on the phone, or simply just a website viewable on all screens
The users can then put up bounties for un-explored areas, and these bounties are bullshit points, they're real money and people can add to the pot.
*not bullshit points
That way, I could do what I'm trying to do now with various random articles researching stuff more efficiently
Since it's edited by the community, it could become a standard reference
crushing MSE into oblivion, sorry guys!
J/k they would be sister sites

2 hours later…
4:29 AM
0

Ultra-modular representative of rational numbers Objective Given a rational number, output the representative in the corresponding coset in the additive (abelian) quotient group \$\mathbb{Q} / \mathbb{Z}[1/2]\$. \$\mathbb{Q}\$ is the additive group of rational numbers, and \$\mathbb{Z}[1/2]\$ is ...

Help. I cannot explain quotient groups to laymen.

4:41 AM
@DannyuNDos huh?

@EM4 Yes I do

2 hours later…
7:02 AM
@copper.hat where would you rank this fight all-time? (can't wait for October :-)

7:49 AM
I prefer definitive results...

8:15 AM
noone can dispute a knockout; but, Fury took it from Klitschko the same way

3 hours later…
10:47 AM
hi

11:01 AM
sup

11:38 AM
Given a measure space $(X,\mathcal A,\mu)$ and $g$ a nonnegative measurable function, then for every $A\in\mathcal{A}$, set $$\nu(A)=\int_A g\,d\mu.\tag1$$ I guess $\nu$ is called the measure of density $g$ with respect to $\mu$. Moreover, for every nonnegative measurable function $f$, apparently $$\int f \, d\nu=\int fg\, d\mu.\tag2$$ I wonder two things; (a) are the integrals in (2) with respect to the same space, i.e. $X$ say? (b) does $f$ have to be nonnegative for (2) to hold?

12:10 PM
a) yes, both measures are defined on $X$
b) I don't think so

ah yeah, ok, thanks. I guess nonnegative is all one cares about in a proof.

1:01 PM

yup

1:40 PM
@psie both measures are defined on $\mathcal{A}$/defined on $(X, \mathcal{A})$ to be pedantic
the domain of a measure is $\mathcal{A}$, and if we want to talk about $X$ we want to talk about the measurable space $X$, not the set $X$ i.e. pair $(X, \mathcal{A})$

1:56 PM
im back
$\int_Axye^{xz} \ \ dxdydz, \ A=[0,2] \times [1,3] \times [0,1]$
@SineoftheTime can i use the trick here?

@Pizza no
it works when you have $\int_A xdV$, $\int_A ydV$ or $\int_A z dV$
the integral $\int_A xye^{xz}$ is not related to the computation of the center of mass

if it was just $xy$? do I have to have like a sum??
I asked you because I had a rectangle, I thought it was possible

2:13 PM
@Pizza nope it does not work

@SineoftheTime okok so this needs to be calculated directly

yes
it's not hard

yes anyway, yesterday I got stuck here, could you help me
$e^{-\sqrt{x}}y = -e^{-\sqrt{x}}(x+2\sqrt{x}+2)+C$

@Jakobian ok :) the reason I'm studying densities now is of course I'm trying to derive the formula of the expectation of an (absolutely) continuous random variable. That is, how we go from $E[X]=\int_\Omega X\,dP$ to $$E[X]=\int_{\mathbb{R}} x\,dP_X(x)=\int_{\mathbb{R}} x f(x)\,d\mu(x),\tag3$$ where $\mu$ is Lebesgue measure.

the differential equation?

2:17 PM
yes

I think this should simply follow from $(2)$ above and the change of variables formula, but I think it is important we allow $f$ in $(2)$ to be an arbitrary real-valued function, since $x$ in $(3)$ is possibly of arbitrary sign.

\begin{cases}
2y' - \frac{y}{\sqrt{x}} = \sqrt{x} \\
y(1) = 0
\end{cases}
this is from the start

ok, why are you stuck?

I tried multiplying both sides by $e^{\sqrt{x}}$

ok, and then?

2:21 PM
wait
I got $C = -5$

you can check on wolfram if the result is correct

solve 2*y'[x] - y[x]/sqrt(x) = sqrt(x), y[1] = 0
can i start wolfram from here so you can see?
i think no

...
$y = e^{\sqrt{x}}x + 2e^{\sqrt{x}}\sqrt{x}+2e^{\sqrt{x}}-5e^{\sqrt{x}}$
i got this
I think I'll do it again from the start
@SineoftheTime being linear of first order, should I apply this formula $e^{-F(x)} \cdot \left(\int e^{F(x)} \cdot g(x) dx\right)$?

2:45 PM
I don't remember the formula
but if it's from your notes, it should be correct

@SineoftheTime how would you have done it?

or using the formula or with the general method

Hey everyone! Recently, I was studying Calculus -3 and am having a real tough time understanding a problem on surface integrals. The problem is from Paul's notes. Can somebody please help me with that? I am posting that in here:
I don't get why they consider the equation of the surface $S_3$ , i.e the disk of radius $\sqrt 3$ to be $x^2+y^2=3$.
This is not the equation of the surface $S_3$ and in fact, it seems to me that $S_3$ does not have an algebraic equation

@SineoftheTime could i use the integral formula?

if it's correct, yes

2:56 PM
Neither I can't seem to make any vector equations, or parametric equations for it.
Am I missing something?
They are pretty much using the formula:

@SineoftheTime I get the same result as before
What do you mean by general method?

does it match with wolfram?

no...

@Pizza this looks correct

But the problem lies with the fact that they consider the surface $S_3$ that should have an eqn $z=g(x,y)$ (with ref to the last picture above) to be $x^2+y^2-3=0$ and this really not what the surface $S_3$ is! It's just the boundary of the surface.

3:02 PM
@SineoftheTime From there I multiplied by $e^{\sqrt{x}}$

@robjohn I need your help :D You are a calculus master.

ok and then?

@SineoftheTime wait
maybe I solved it
$y = -x-2\sqrt{x} -2 +Ce^{\sqrt{x}}$
$0 = -5+Ce$
$C = \frac{5}{e}$
$y = -x-2\sqrt{x} -2 +5e^{\sqrt{x}-1}$
yes

@ThomasFinley it's not written anywhere that $g(x,y)=x^2+y^2-3$
In fact here $g(x,y)=0$
where $(x,y)\in D$
@Pizza does it work now?

yes
Was there a faster way to solve it?

3:20 PM
I don't think so
using the formula is straightforward

@SineoftheTime $\frac{dy}{dx} + P(x)y = Q(x)$
this one?

this is not a formula
this is the general form of a first order linear equation

I don't understand what formula you are referring to

there's a formula to solve this kind of equations
you can search it online

@SineoftheTime Ok, that makes sense. Now, I understand what they meant. Thanks for the clarification!

3:29 PM
np

@SineoftheTime I could also go in this form, right?
$y' = \frac{\sqrt{x} + \frac{y}{\sqrt{x}}}{2}$

I don't see how this is useful
you should study the theory before doing exercises

@SineoftheTime ok

4:15 PM
@SineoftheTime thanks , now its more clear

4:28 PM

1 hour later…
5:37 PM
@DanielDonnelly you may try discord?

For real sequences, absolute convergence does not imply convergence right?

ye

gracias

no wait

series or sequences?

5:51 PM
6

Let $V$ be a normed space with norm $\|\|$. $\sum_{n=1}^\infty a_n$ is an absolutely convergent series if $\sum_{n=1}^\infty \|a_n\|$ converges. Could you please explain why in complete normed spaces the absolute convergence implies convergence, but it doesn't hold for incomplete normed spaces?

$\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$
how to latex

sequences
for series i believe it's true

I thought sequences were just finite series

@EE18 what do you mean by abs. convergence of a sequence?

That |x_n| converges

ok, take 1,-1,1,-1,...

6:02 PM
Ya you're right, this was a silly q :)
i was getting confused about series and the reult being true there

6:15 PM
Oh wait
You wrote absolute convergence implies convergence, then yes for series

Another question about sequences. Boring, I know, but I want to briefly be able to manipulate this easy stuff before going on
Obviously we have $(n+1)(n+3) \to 1$
Other than getting bogged down in $\epsilon$ stuff, how would one show this?

The Rouché-Capelli theorem states that there exist solutions for the system if and only if the rank of the augmented matrix is equal to the rank of the coefficient matrix. So if I calculate the determinant and I get $det = 0$, the system can have infinite solutions or no solutions, but since I found solutions exist, are they infinite?

\epsilon stuff is the definition of the limit. There's no escaping it.
@EE18 you would use something called 'limit rules' but those also come from the \epsilon stuff.

6:31 PM
@EE18 it does?
As $n$ does what?

@Pizza yes. There are not many possibilities for a linear system of equations: 1) no solution, 2) unique solution, 3) infinitely many solutions.
@XanderHenderson they most likely made a typo there. I think they meant (n+1)/(n+3).

or n tends to sqrt2 - 2

EE18: limit theorems. the calculus trick of rewriting the thing as (1 + 1/n)/(3 + 1/n) is putting that sequence in a situation where the usual limit theorems are clearly applicable to it (i.e., ratio of sequences that themselves tend to real limits, perhaps also via limit theorems)

@XanderHenderson they also said sequences :)

@Koro I mean, is there a need to calculate the determinant, if for rouche capelli I already know that there is a solution?

6:37 PM
if you had to deal with it from the definition (which literally nobody outside of the first week of a class should ever do), the simplest way would be to use polynomial division to rewrite as 1 - 2/(n+3) because the algebra associated with choosing an N for the epsilon is easier with only a single n in a denominator (indeed, even the smallest N for a given epsilon, which literally nobody should be trying to find ever, is easy to locate with this rewriting)

@Koro not in a context which made it obvious that this was the intended context. And since $(n+1)(n+3)$ clearly doesn't go to 1, clearing up the confusion would be good.

EE18: many of the 'rewriting tricks' that people absorb by osmosis have roots in making it clear that a given example can be handled as a special case by [some toolkit of known theorems], with the logic of the trick usually motivated by the exact contents of the toolkit [which can be highly textbook specific], and vanishing once there is no exact toolkit

@Pizza yes. If it is known that there is a solution, then one wants to know if it is unique or not. If the determinant is non zero, then one gets a unique solution.

@Koro if I get 0, but I know that there is a solution, then are they infinite?

i feel like the list of names in en.wikipedia.org/wiki/Rouch%C3%A9%E2%80%93Capelli_theorem needs an additional bullet point which says "also, in many of these countries, it might be that the majority of the people familiar with this theorem do not know it by any name whatsoever"

6:50 PM
@leslietownes haha

good example of how references maybe don't give the most accurate picture of usage, haha

I learnt the name today myself. :D

Same here, I never expected it to have a name

or maybe some bullet points "also known as 'Theorem 3.16' in classes taught out of X book" "also known as 'Theorem 7' in classes taught out of Y book" ...
i guess the general problem like this is that (a) for inclusion in any kind of organized reference, it does need some kind of canonical name, (b) an assertion like "by the way, literally nobody calls it this" is difficult for a wikipedian to back up with citation

fame distribution

6:57 PM
i like how in the list of bullet points, it's pretty clear that in some countries there's an effort to shoehorn a native of that country into the name of the theorem
i'm surprised we don't try to call it the sylvester-rouche theorem here and in england

and earlier concepts/results were given original names but now they are just given generic names like - good function, bad function etc.

I'll call it Ramanujan's theorem from now on

@Koro can you check my last msg pls

making it difficult to find a reference on them online.
@Pizza that's right.

@Koro 👍

7:01 PM
oh you mean the Pushkin-Tolstoy theorem?

@ThomasFinley calculus -3? So you're going backwards or something?

haha

he lives in $\Bbb Z/6\Bbb Z$

> The teacher spoke to Danny: "Why are you on the floor?"
Danny said: "Because you said to do this Math problem without Tables."

7:22 PM
what is calculus 1,2,3 etc.?

koro: course names? what they "are" are depends on the school
often, "calc 1" is some version of computation with limits/derivatives, maybe an intro to integration, "calc 2" is integration, maybe with series, and "calc 3" is maybe multivariable calculus
but not always

Thanks on all fronts to everyone's comments above, very helpful for me

also, haha, what is "multivariable calculus"
do you stuff differential equations into any of that, and if so, where
these are the many variables

in the west calculus is subdivided into 3 (sometimes 4) parts where the first part introduces limits, derivatives, sums and integrals. The second part typically continues with integration techniques, moving onto series, series expansions, and some other topics I forget :P 3 is just multivariable calculus

yeah, makes sense. I have seen lot of people online asking- I have a question from calc 3 etc. making it look as if it is universal.

7:25 PM
ah yes, the renowned curriculum followed by "the west"
i don't normally use ":P" but would consider it here

No, calc 1/2 might have veeerry elementary diff eq but 3 usually doesn't (from what I've seen)

I can say I have a doubt from math 101 :)

koro: because students about math seem to assume that their math environments are universal (as people on the internet more generally assume about their experiences with anything)

2

I see a lot of places where "Calculus 1" is referred to as "Introduction to Calculus", or "Single-variable Calculus." "Calculus 3" is referred to as "Multiple-variable calculus." Is there an alternate name for "Calculus 2"? Am I mistaken in thinking single variable calculus is Calculus 1?

My own answer to that question: matheducators.stackexchange.com/a/16919/8571

it's weird, i think most people realize if they are complaining on the internet about their experience in buying a car, or something, that they aren't referring to a particularly specific experience
but math being objective and having this component of being less culturally dependent than other things causes people to forget that when it comes to things like course titles, education systems, how things are taught, when people learn what things
and there is a limited amount of commonality across a range of experiences of "calc 1" "calc 2" type categories, at least within US universities/colleges/(god, how you name those is another one of these), which doesn't help

7:30 PM
I think those people are pretty ignorant with math as a subject though so it makes sense that all they know is their curriculum and assume that everyone can infer what they mean
The benefit of MSE is that at the end of the day the question has to be intelligible at least

like yes, in america we all generally drive on the right hand side of two-laned roads and gripe about the same general kinds of driver behavior
even though traffic law varies among the 50 states

I don't. I drive in the middle of both lanes

and stuff like where it's OK to park can vary within a city depending on the day and time

@Obliv haha. why?

In my lifted pickup truck with highbeams on blast and a beer on my dash /s

7:37 PM
I'm meant to use the binomial theorem to show that $n^3/2^n$ is a null sequence, and I've technically done it, but it really is the most hideious proof I've ever displayed. Almost embarrassed to show ithere, but basically boiled down to a guess and check on an ugly base case and then a basic induction argument. Is there a smarter way to do this?

Hmm, perhaps I've not seen that yet in my book
will look that up

can't you just use the hospital rule

This is one of those ones where Leslie will rightly point out that the answer depends on what toolkit I have
Suffice it to say that my toolkit here is minimal. It includes basically only the "comparison test"/"squeeze theorem", that for real sequences if my sequence is between two which converge to the same limit, then so too does mine

You know squeeze principle?

7:41 PM
Yes indeed :)
I sort of use it in my argument pictured above
But the argument is hideous given that I guess and checked a base case at $n = 16$

why can't u just take the derivatives of top and bottom

So in your sequence (x_n), lim $(\frac{x_{n+1}}{x_n})=1/2$

Ah that's the ratio test. Haven't seen that one yet but will make a note that it's the way to do it. Maybe with my current toolkit my awful argument is the only way to go

@EE18 Jeez... that seems crazy convoluted.

$\lim_{n\to \infty} \frac{n^3}{2^n}$ becomes $\lim_{n\to \infty} \frac{3n^2}{2^n\ln2}$ yada yada until you have 0 on the top
or are u not allowed to take derivatives

7:46 PM
Koro, it seems like ratio test is for series rather than sequences

Use limit definition now to say: $0 \le x_{n+1}\le 3/4x_n$ for all large n.

Ya Obliv, no derivatives for at least two more chapters :)

@Obliv "The Hospital Rule"? Are you trying to make a joke?

From here, note that $x_{n+1}\le (3/4)^2 x_{n-1}$

7:47 PM
L'Hospital translates to The hospital

@Obliv That isn't "the Hospital". It is a person's name.
It doesn't "translate" as "the hospital". :/

repeat this step to get: $x_{n+1}\le (3/4)^n x_1$.
Now, use the squeeze principle.

sometimes spelled l'hospital

@Obliv Yes, I am aware.

I know it looks like a joke but it's the same if I said use "the euler rule"

7:49 PM
@EE18 this only assumes that $\lim_n a^n =0$ if a is in the interval [0,1) and the squeeze principle.

@Obliv That, also, would be wrong.
(In most dialects of English, anyway).

Oh wait I see the issue. His name is L'hospital so it'd actually be "the L'hospital" rule

@Obliv That, too. But "the L'Hospitals rule" sounds awkward. Noöne speaks like that.

@EE18 no, it is for sequences too. But anyways, forget that I used the term 'ratio test'.

7:53 PM
"L'Hospital's rule" is the name of the result in the English speaking rule. I think that people will generally understand you if you say "the L'Hospital rule", but it sounds strange (to me, at any rate).

yes I think "the L'hospital rule" sounds better than "the L'hospitals rule"

@Obliv Yeah, that isn't the issue. The result is not called "the [anything] rule". It is "L'Hospital's rule". There is no article in front.

Why are we talking L+ ratio suddenly?

@EE18 yw! Although, when you learn 'ratio test', you'll see that what I did above is how the test statement is proven.

does it matter if [anything] is a person or something else? I wonder if it works better when it's not a person in the brackets

7:55 PM
@Obliv I am not making a general statement. I am specifically saying that, in English, the result should be called "L'Hospital's rule", not "the L'Hospital rule".
"L'Hospital" is not an adjective which modifies "rule". It is possessive (genitive)---it is the rule which is possessed by L'Hospital. ("Possessed" in the sense that he is given credit for it; though it was actually a Bernoulli who proved it, but L'Hospital had the money, so got to slap his name on it).
@Koro I just said that.

If only I were good at english :P
So instead of "the Lagrange theorem" we say "lagrange's theorem" etc

@XanderHenderson I didn't see that somehow. Or probably, I posted it as soon as you did.
:)

@Obliv Yes.
(In that case.)
There are cases where the name has become an adjective, e.g. it is more correct to say "the Pythagorean theorem" than "Pythagoras' theorem". So I wouldn't try to infer a general rule, here.

Seems legit.

8:43 PM
Am in the process of proving this but want to ask re: terminology (so DKM)
Is it fair to say that "this claim says that convergent sequences are Cesaro summable and that their Cesaro mean equals their limit"? i.e. we give a sufficient (but AFAIK not necessary) condition for Cesaro summability
In mathematical analysis, Cesàro summation (also known as the Cesàro mean or Cesàro limit) assigns values to some infinite sums that are not necessarily convergent in the usual sense. The Cesàro sum is defined as the limit, as n tends to infinity, of the sequence of arithmetic means of the first n partial sums of the series. This special case of a matrix summability method is named for the Italian analyst Ernesto Cesàro (1859–1906). The term summation can be misleading, as some statements and proofs regarding Cesàro summation can be said to implicate the Eilenberg–Mazur swindle. For example, it...

@EE18 yes
its correct to say that

Is this an interesting notion at all?
I'
ve never heard of it before

or perhaps it'd be more correct to say its Cesaro convergent
Cesaro summable is more about series

"Cesaro summable" is perfectly correct.

yeah actually I think you shouldn't use Cesaro summable in this context

8:51 PM
@Jakobian Why not? It is completely correct here...

a series $\sum a_n$ is Cesaro summable when the sequence $s_n = \sum_{k=1}^n a_n$ is Cesaro convergent
where by Cesaro convergent I mean $\lim_{n\to\infty} \frac{s_1+...+s_n}{n}$ exists
saying that $a_n$ is Cesaro summable might be confused for above condition for $\sum a_n$

I disagree. The sequence $\{a_n\}$ is summable if and only if the series $\sum a_n$ converges.

FWIW, this is what I wrote in my notes (I think I see what Jakobian means now):

this is not equivalent to $a_n$ being Cesaro convergent

"Summable" is a description of a sequence.

8:53 PM
Can you reread the conversation Xander

Ah I see what Xander means now too lol
What I wrote is nonsense

@Jakobian I did, and the terminology looks perfectly fine to me.

Which terminology? The one EE18 is using?
So you disagree with my terminology?

12 mins ago, by EE18
Is it fair to say that "this claim says that convergent sequences are Cesaro summable and that their Cesaro mean equals their limit"? i.e. we give a sufficient (but AFAIK not necessary) condition for Cesaro summability
Sequences are summable. Series have limits. (Sequences also have limits, but that isn't the contrast I am trying to point out.)

Here Cesaro summable is what I call Cesaro convergent
$(x_n)$ is convergent implies that $(x_n)$ is Cesaro convergent in my terminology
$(x_n)$ is summable implies $(x_n)$ is Cesaro summable in my terminology

8:59 PM
What is your definition of "Cesaro convergent"?

7 mins ago, by Jakobian
where by Cesaro convergent I mean $\lim_{n\to\infty} \frac{s_1+...+s_n}{n}$ exists
This is definition of $(s_n)$ being Cesaro convergent

@Jakobian That is the definition of Cesaro summable...

that is definition of $(a_n)$ being Cesaro summable according to my terminology
as well as wikipedia
where $s_n$ are partial sums of $a_n$

You seem to have said that a series $\sum a_n$ is Cesaro summable if the corresponding sequence of partial sums is "Cesaro convergent".

that's right

9:02 PM
But a series is not any kind of summable. Sequences are summable. The sequence $\{a_n\}$ is Cesaro summable if the sequence $s_k = \sum_{j=1}^{k} a_j$ is convergent.

I don't distinguish between "$\sum a_n$ is summable" and $\sum a_n$ is convergent"
well that's irrelevant for the purpose of conversation but we can disagree on that

@Jakobian That is bad English. A sequence $\{a_n\}$ can be summable or not; it is summable if the corresponding sequence of partial sums is convergent.

At the risk of interjecting, may I ask again:
Is this an interesting notion at all? I've never heard of it before so not sure

@EE18 Is what an interesting notion?
Cesaro summability?

Cesaro summability

9:03 PM
Sure.

what I'm saying is that EE18 said above is not what people call Cesaro summability of $(x_n)$ but Cesaro convergence of $(x_n)$
Cesaro summability comes with extra step of taking partial sums

@EE18 See Stolz–Cesàro theorem which is quite useful

Very cool, thanks for the heads up Soumik :)

and as for whetever we should call $\sum a_n$ summable when it is convergent, this is terminology that e.g. Hardy uses
in his book Divergent series

9:09 PM
As I read it, the original question was "Is it reasonable to call the sequence $\{a_n\}$ Cesaro summable if $\lim_{k\to\infty} \frac{1}{k} \sum_{j=1}^{k} a_j$ exists?
The answer to that question is "Yes, this is perfectly correct."

According to you but not to e.g. wikipedia

I have certainly seen authors describe series as summable, e.g. "$\sum a_n$ is summable if...", but that is a kind of loose shorthand for.

Do you agree that the sequence $(a_n)$ is summable when $\sum a_n$ is convergent?

@Jakobian I think that you think that I am saying something stronger than what I have actually said...

that those two things mean the same

9:12 PM
And chat is not working properly...
The phrase "the sequence $\{a_n\}$ is summable if..." has meaning, and is perfectly correct. That doesn't mean that other phrases are not also understood. It just means that the phrase I've written above is perfectly fine.

What is this meaning according to you
I don't understand where this disagreement is coming from
I myself used to confuse $\lim_{n\to\infty} \frac{a_1+...+a_n}{n}$ by Cesaro summability of $(a_n)$
but its not what $(a_n)$ being Cesaro summable means
the Cesaro sum of series $\sum a_n$ is equal to $\lim_{n\to\infty} \frac{\sum_{k=1}^1 a_k + \sum_{k=1}^2 a_k + ... + \sum_{k=1}^n a_k}{n}$
not to $\lim_{n\to\infty}\frac{a_1+a_2+...+a_n}{n}$

You seem to have implied that one should not call a sequence [Cesaro] summable. But "summable" is an adjective which applies to a sequence. Calling a series any kind of summable is done, but is slightly loose.

this is because we want convergent series to be Cesaro convergent to the same thing
No, I said we should not say that $\lim_{n\to\infty} \frac{a_1+...+a_n}{n}$ exists to mean that $(a_n)$ is Cesaro summable
at least that's what I meant

2 hours later…
11:33 PM