5:12 PM
thanks matt

no problem : )

"Actually I don't need to understand the proof, I just need the proof itself"...

since you want to study (u1+u2+...+un), you should write it as (u1+u3+...+u(2m+1)) + (u2+u4+...+u(2m')) and study each sum separately.
@HenningMakholm that screams homework problem

Is there actually a difference if I write the series (-1)^(n)/(n-n^(1/2)) as (-1)^(n+1)/(n-n^(1/2))? The only difference is the coefficient on -1, instead of n being n+1... it shouldn't make a difference right? Wolframalpha gives me different results.

@mercio no kidding.

5:22 PM
I commented and voted to close.

@Clash: this changes your sum by more than 2/(2 - sqrt(2)) because the first term in your sum (for n=2) is 1/(2-sqrt(2)) in one case and -1/(2-sqrt(2)) in the other case. What do you think, @Srivatsan?

I must print that and put it on my office door.

: )

@Matt A finite prefix of the series can be cut out without guilt, no?

@Matt I just wanted to test the convergence, this shouldn't affect that, right?
gets me converges for n and diverges for n+1

5:26 PM
@Srivatsan: I'm not sure what you mean. If you sum 1/2^i it makes a difference whether you start at 1 or at 0...

@Matt It matters to the sum, not whether you can sum it.

@HenningMakholm I believe it might be a language problem. Perhaps he wanted to say: I need also the proof, not only the explanation. And he does not speak English good enough.
We have here many posts that have language problems.

@Matt To put it less rhetorically, it affects the sum but not the convergence. Clash is interested in the latter.

@Srivatsan: That's what I meant. : ) Now I understand: @Clash just wants to know whether it converges or not : )
Should've read your conversation from the beginning before barging in on it. But I was distracted by p-names and forcing posets...

@AsafKaragila I expected that to be closed, but as a duplicate. (I am surprised that proof is not already at this iste.)

5:30 PM
Perhaps.

@Matt so, it shouldn't affect convergence, right?

@MartinSleziak You should find the duplicate, then.

@Clash: I don't see how it can since you can move -1 in front of your sum. I think I'm missing something, let me think.

@AsafKaragila I'm trying to. As I said, it will be surprising, if there is none... but still it's possible.

@Matt i dont think you got it wrong, wolframalpha is probably crazy

5:35 PM
@Clash: That just seems so improbable...

by the way, n+2 and n+3 also diverge... apparently it converges only with the coefficient n

@Clash: Precision error in the code that does the computation maybe?

@Matt could be, I don't really know how it works, but I trust more you than it

@Clash: You could file a bug report.

Sorry, @Clash, I missed the discussion while I was posting an answer.

5:41 PM
@AsafKaragila The only thing I was able to find was this one: math.stackexchange.com/questions/38900/binomial-expansion It's not a duplicate, since it asks about a specific method of proving this identity. Either I should try to improve my searching skills, or it is really not on math.SE.

@Srivatsan no worries mate, I ended up posting the question, but thank you very much for the help!

@Clash Oh, that post was from you =).
I just opened the question in the other tab.

yeah =) but thanks a lot for the insight, it was very helpful!

You're welcome.

hmmm

5:51 PM
@Clash: Martin's answer is perhaps the clearest and easiest to write as a proof; Fezvez's answer matches my intuition more clearly. In the long run, I think Fezvez's approach is easier to see what's going on. I suggest you understand the approach if not precise estimates used.
[For completeness sake, let me add that I don't understand Davide's answer is coming from. =)]

oh I have three answers now, I have to study fezvezs reply
davide did make a comment about his reply, but I didnt understand it

Oh, it might be a good answer. I just find it too clever.

i had never seen this Euler–Mascheroni constant thing before

Well, it is quite irrelevant. But you will need to know that H_n = log n + O(1).
If you understand what that means...

is O(1) the big O notation? meaning O(1) is just a constant?

5:58 PM
Closing that binomial question was really interesting: 3 votes for too localized, 1 not constructive, 1 exact duplicate...

Of course, the too localised gang is right. ;)
[No points for guessing why.]

@Martin, @Henning: You could ping Bill and tell him to see above :-)

@Clash: Yes.

@Clash Yes, it means that H_n never deviates from log n by more than a constant.
This means, there is some constant A such that log n - A <= H_n <= log n + A for all n.

thanks matt and srivatsan for the explanations!

6:12 PM
No need to thank me.

You're welcome, Clash.

when calculating this limit: lim n->infinity (-1)^(n+3)*(n+2)/(5n^2+1)^(1/2), can I remove the alternating part? and say that it goes to result * -1 or result * 1? For that case, 1/sqrt(5), so 1/sqrt(5) or -1/sqrt(5)
therefore the series diverges, as lim n -> infinity != 0?

well a series whose term doesn't converge to 0 can't possibly converge.

which term, the "last" one, right?

the nth term, I think.

6:20 PM
urhg
I commented on someones answer "that's wrong because .." then he said something and I said "oh my mistake I thought you were talking about X but you really meant Y"
now they trying hard to find mistakes in my answer ...

@Clash I don't think an infinite sequence has a "last" term

@QED Do you want me to pitch in? =)

well

I have added a comment which hopefully clears it up
4

First note that $$\lim_{x \to \tfrac{\pi}{2}^-} \sin(x) = 1$$ and $$\lim_{x \to \tfrac{\pi}{2}^-} \cos(x) = 0.$$ therefore $$\lim_{x \to \tfrac{\pi}{2}^-} \frac{\sin(x)}{\cos(x)} = + \infty$$ since we have (by the definition of limit): \forall \varepsilon_1, \exists \delta_1, \forall x < \...

before I was complaining about math.stackexchange.com/q/85806/16697 being wrong
in fact it's not
it's just horribly confusing because "lim ... = infinity" actually has nothing to do with equality
so it's negation has nothing to do with "not being equal to infinity"

6:25 PM
Ah alright.. Same thing I pointed out yesterday. Just that TonyK is a little more persistent. =)

so I got distracted with all that notational rubbish, when all he was really saying is that we should be considering one sided limits (which is correct)

I was confused because it looked like you commented on your own answer saying that you mistook yourself and as a result, other people tried to disprove your answer

The way I wrote it was: A and B therefore C -- <definition of A> and <definition of B> therefore <definition of C> by [proof method]: verification: showing that proof method actually works here

@QED True that. That's a subtle point...

@mercio, I was commenting on someone elses answer but I aknowledged that I was mistaken and deleted it

6:27 PM
@QED yeah and then someone argued your answer, which is a different answer than the on you commented one

yeah

your original line was unclear, that's all

what line?

so it looked funny to me
@QED this one

oh right
It seems very difficult to write proofs in a clear way that other people can understand

6:29 PM
guys is there any sequence a_n, where lim n -> infinity a_n = 0 and \sum (-1)^n*a_n diverges? I wanted to use something like a_n=2^(1/n), but that clearly isn't going to get me lim = 0...

I wanted to emphasis the overall structure then show the important details, but he seems to be getting lost reading it

@Clash Let's start with a_n = (-1)^n b_n, where b_n is the harmonic series. Wouldn't quite work, but you can fix it.

alright, sum will diverge, but the limit isn't 0... to make it 0 we can divide by n, I guess
awesomeeeeee and then we get 1/n, yay, which diverges

@Clash wait, dude! =)

a_n= (-1)^n/n, oh I still can't say limit = 0, or can I?

6:32 PM
Let's see what it gives. (Condition 2) sum (-1)^n a_n = harmonic series, diverges. Check.
Condition 1: Is the sum zero? The sum is log 2 or - log 2 or something like that. Not zero.

Which sum should be zero? Just the limit from a_n should be zero, or did I get something wrong?

Let's say that the sum is S (that I don't care about). Define the new series A_n = a_n for n > 1.
@Clash Er, you're right. I got it wrong. That example itself might work.

@Clash try an = 1/n for even n, and an = 0 for odd n

whcih sum were you thinking about? didn't really understand where you got the logs
a_n = (-1)^n / n -> lim = 0, and sum diverges, everything checked, right?

@Clash Sorry, I was thinking of the sum of the series sum a_n, which is the alternating harmonic series.

6:35 PM
oh I see, but yeah, the exercise only says limit

Why did I think that? Because I don't read the question properly, that's why. =)

thanks for all your help!! I am now done, but really scared with the exercises this time! They proved to be easier than I thought
oh it happens srivatsan, I can't imagine how many question you have probably read today

I mean with your guys help of course it ended up to be easier ;)
but really, some proofs this time were really short, I always think I did something wrong when it's small

6:37 PM
sadly no :( but I'll try, lol. jk

The material is organised to keep any individual proof short. That's what lemmas are for.
@QED Don't get disheartened. I think TonyK has a point. Let's first see what that is...

yeah but the other exercises weren't that short/easy to proove
for example, I have the series k^2011/(3^k), if you do the root test, you get straight off 1/3, which shows it converges

Um, are you sure you are doing them optimally?
Ratio test works as handsomely, no?

then we have the series k!/k^k, we know that k! < k^k, then if you do root test, you know it's < 1, which means it converges
yeah I guess radio test works too, but <3 root test, so easy

what's the root test ?

6:41 PM
sec, lemme get you a link
In mathematics, the root test is a criterion for the convergence (a convergence test) of an infinite series. It depends on the quantity :\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}, where a_n are the terms of the series, and states that the series converges absolutely if this quantity is less than unity but diverges if it is greater than unity. It is particularly useful in connection with power series. The test The root test was developed first by Augustin-Louis Cauchy and so is sometimes known as the Cauchy root test or Cauchy's radical test. For a series :\sum_{n=1}^\infty a_n. the...

@Srivatsan, I think he's just randomly objecting to things because I commented on his answer :/
If there is a mistake in the proof I'd like it pointed out but.. there isn't...

@Srivatsan but I haven't done anything wrong this time, right? It's because on the previous exercises we could only use the comparison and the alternate series test. and that's more stuff to write

@QED Maybe, but you gave him a chance to. And it's not like anything is lost, right? You can rectify the mistake, improve the answer? At the least you can ignore the comments =)

@Clash you can only use two things ? you can't use basic logic ?

for the alternate series test, we have to show lim = 0, monotone falling and a_k > a_k+1... for the comparison test, well you have to compare it, that's more stuff to write
@mercio what do you mean with basic logic? I just mean we couldn't use for example the ratio test, because we hadn't learned it... if by logic you could reach that test, you still wouldn't be allowed... and yes I also disagree with this, but well, im not the professor
and i dont make the rules

6:44 PM
@Clash Can you elaborate the answer for this for me?

@Srivatsan the answer to which question? sorry

First, does this series converge or diverge? Next, how to show that?
[Click the arrow to the left.]

@Srivatsan oh ok, well, we know that k! < k^k. we know that the root test (k-th root) for k^k is k, and that the root test for k! is smaller than k. then we have something smaller than k divided by k => it's smaller than 1 => it converges by the root test
is this proof enough?

what do you mean, you know that the root test for k! is smaller than k?

@Clash :(a.) It does seem to converge, but the proof is wrong (not just incomplete). (b.) Ratio test works out to be easier than root test, I think.

6:49 PM
converge?
oh you mean Clash

Oops, sorry.

@mercio I mean the k-th root from k! is smaller than k

I was worried for a sec :P

Sorry for startling you... =)

@Clash: look at this way: 5! = 5*4*3*2*1 < 5*5*5*5*5 = 5^5

6:51 PM
@Clash so the kth root of uk is less than 1, so what ?

@Clash That ain't enough. What does the root test give for the harmonic series? For the series sum 1/n^2? The test is that the nth root of a_n should be bounded away from 1. That is, it should be at most some k (independent of n) that is strictly smaller than 1.

@QED yeah this is exactly what I mean, 5!^1/5 < 5, and then we have something smaller than 5 / 5, which is smaller than 1. by the root test it converges then

I was only talking about n! < n^n (for n >= 3)

Is it possible to link to a comment?

@Srivatsan it tells me nothing, it just says = 1

6:52 PM
That's the same problem with n!/ n^n, no? Well, actually, the root test does work, but you should use better upper bound on n!.

to prove it by induction you need to generalize it so that you're really proving n!/(n-k)! < n^k

@Srivatsan for 1/n^2 we get 1/((n)^(1/n))^2, which is 1, and therefore we can't say anything about that serie
or am i applying the root test wrong?

@Clash no I'm afraid the root test is "if nth root of un < a < 1 for all n, then the series converge" and not "if nth root of un < 1 for all n, then the series converge", which is false because the harmonic series diverges for example
where a is any fixed number

@Clash No, it seems that you got the root test correctly. I don't understand what you are doing for n!/n^n though.

@mercio but the root test doesn't say anything about the harmonic series, you get 1, it needs to be smaller than 1 to say it converge

6:55 PM
@Clash but nth root of 1/n IS smaller than 1
how do you edit these things

Press uparrow and you will allowed to edit the previous comment. There's a 2 (or so) minute time window before which to edit. @mercio

@Srivatsan well n! < n^n, so (n!)^(1/n) < n, so it converges
@mercio ? no it's not? 1^(1/n) = 1, n^(1/n) = 1, 1/1 = 1

@Srivatsan thanks

@Clash Incorrect. If a_n = n!/n^n, then it is true that a_n^{1/n} < 1, but that is not enough. (*Think* more carefully about the harmonic series example.)

@Clash what do you mean, n^(1/n) = 1 ? I'm quite certain 2^(1/2) > 1

6:58 PM
@mercio but n->infinity

@Clash alright, so how do you know that nth root of (n!/n^n) doesn't converge to 1 ?
you only know it's less than 1 for any n
(just like n th root of 1/n)

@mercio I don't know to how much it converges, I know it's less than 1 and that's all I need for the root test? that's how I understood the test
@mercio no, n-th root of 1/n is 1, it's not less than 1

no, the limit of nth root of 1/n is 1, but nth root of 1/n is less than 1, so it's not 1

@Clash Of course, that's nonsense. The nth root of 1/n is strictly less than 1. What is 1 is the limit as n goes to infty.

oh sorry, yeah, I ment the limit, and that's all that matters for the test

7:00 PM
likewise, the nth root of n!/n^n is less than 1, but you have to show that it doesn't converge to 1
(and that's not even sufficient)

I still haven't understood why I cant use the root test :( and I'm pretty sure you guys know it better than me
I don't see where I went wrong, the limit is < 1 and then the test says it's convergent

well, what's a precise correct statement of the root test ?

upload.wikimedia.org/wikipedia/en/math/3/e/2/… < 1, the series converge, > 1 series converge

@Clash Right there. Why do you know that the limit is < 1? You only showed that each individual term is smaller than 1...
Example: each term of the sequence 1-1/n is smaller than 1. But the limit is equal to 1, not less than 1. Same with 1/n^{1/n}

@Gigili Yes, but it is not always that easy to get the link - have a look here: chat.stackexchange.com/rooms/36/conversation/… and it is also mentioned in this answer on meta: meta.math.stackexchange.com/q/3239/8297

7:03 PM
@Srivatsan well, even if k-> infinity, k! is still going to be < than k^k, right? this is the only limit that interests me

@Gigili If it is a recent comment, looking at the activity (comments tab) of that user seems to be the simplest method.

@Clash Yes, that shows that n!/(n^n) is less than 1. But as you take the limit, you can only conclude that lim n!/ (n^n) [if it exists at all =)] is less than or equal to 1.

@Srivatsan yeah, my proof is not about saying that if each element is < 1, than it converges. As 1/n diverges and each term is smaller than 1. But the test says, if you take the n-th root and make the limit n->infinity, if it's smaller than 1, it converges. have I done something wrong? really sorry if im repeating myself and didnt understand u guys
ohhhhhh equal to 1
but why equal to 1? I mean k! is always (k>1) smaller than k^k

I am starring that comment =)

lol

7:06 PM
That's a good question. Think about the two simpler examples where it's clear what's going on. (1) 1 - 1/n. (2) 1/n^{1/n}.
If the sequence can get arbitrarily close to 1 (and the limit exists), then the limit has to be 1. No other go. =)
That's the case with 1/n^{1/n} and 1-1/n. True that they are small than 1 individually, but the sequence--as a unit--gets close to 1, and that's all that matters for the limit to be equal to 1.

ohh I see, both sequences tend to 1 when n->infinity
and then you get 1/1, which gets us equal 1 to the end
even though one is always smaller than the other...
well for me it's almost a contradiction, but ok, math, i guess there are some things i dont understand

That doesn't happen with the sequence n!/n^n, but that's not a trivial statement. Needs some proof. [If you are interested, the limit of n!^{1/n} / n is equal to 1/e, which is strictly smaller than 1 and you are done. But establishing this formula is a bit involved.]

My god, it's like Bill was reading the comments in the wrong order but he then insists that he doesn't care. I guess I should have guessed that the OP just wanted to see a proof and understand it on his own... before even saying so.

I guess asking on chat is the easiest way: Can I find 2 more people voting here for reopen - math.stackexchange.com/questions/86093 (And then 5 votes for closing the question will be needed - but this time as exact duplicate, not too localized.)

@Srivatsan awesome explanation srivatsan! really many many many thanks man!

7:11 PM
@Clash No, please don't do that =)
But cash is fine. Do you want me to send my address? =)

lol sorry, just another poor student
but really, nice to meet you man! you are very kind and humble! I really wish I could be more like you
keep on! cya, I'm going to have dinner

See you. Time for my lunch as well.

@AsafKaragila I guess after reopening and closing as an exact duplicate, many comments from that discussion will be obsolete, so we can perhaps delete some of them.
Even if only out of respect to Bill Dubuque - if he likes this solution better - reopening and closing is worth the trouble.

I can't cast another closing vote.

7:15 PM

@MartinSleziak I don't quite understand the routine, but I voted to reopen. =)

ah that was a terrible question
he didn't define the binomial coefficient
I don't think you should re-open this, it would be better to ask it yourself

Some people did not like the reason for closing, so it will be reopen and after that immediately closed - but this time as exact duplicate.

that seems silly

Well, it will have the correct reason for closing.

7:17 PM
We can just delete it altogether.

that's a good idea

- delete on what grounds? =)

due to it sucking lol

@QED If I may quote Michael's comment: _ I've voted to re-open. In the notice explaining that it was closed, it says "This question is unlikely to ever help any future visitors". That doesn't make any sense at all. It seems as if closure resulted from people misunderstanding the poster's comments as meaning he doesn't want to understand the proof but just wants to copy it so he can turn it in. _

7:19 PM
- If that is not the interpretation, what is? I can't make anything out of a person coming and asking me to prove a random identity. It's (1) homework, (2.) he is interested in understanding in which case the OP's comment makes no sense, (3.) he wants to get his own understanding from a formal proof.

As I said the question is terrible because it doesn't define the binomial coefficient

But I agree that this question caused a lot of chaos here.

I reopened the question by casting the last vote.
Now who's gonna close it? Do you have the link to the duplicate?

In any case, I still think that the question is localised. At least till the OP explains themselves and their strange request.

7:23 PM
Actually, one question that has been nagging me: what do I call «that post this question is a duplicate of»? The «original»?

@Srivatsan: Could you please now vote for closing that question, but this time as "exact duplicate"? Thanks a lot.
I guess this was what Michael was hoping for, when he voted for reopening.

Can't vote a second time. Same problem as with Asaf.

Same problem with me :-(

I think I can vote...

@ZhenLin It's this question: math.stackexchange.com/questions/86093
Thanks a lot and sorry for the inconvenience.

7:28 PM
@MartinSleziak Thank you so much.

@Gigili Gern geschehen. Hoffentlich es hilft.

Done.

Thanks Zhen.

Good, although I have a feeling the OP will not take much notice of that

7:31 PM
Hope I don't anger someone.

BTW we will have 1k posts on meta soon.

@Martin I would imagine that you too couldn't vote in the second round, right?

@Srivatsan Right.

Hello, a simple question: Given an interval [0.t], "for partitions Π1 and Π2 there exists Π=Π1∪Π2 which is refinement of both". Why is Π1∪Π2 a partition?
This is from a comment here math.stackexchange.com/questions/85866/…

@Ethan You can understand partition as a set of points from the given interval.

7:40 PM
partition is a set of subsets of the interval. Why is it a set of points?

But if you define partition of [0,t] as a finite sets of intervals
[x1,x2],[x2,x3],...,[x_{n-1},x_n]
Then you see that the partition is uniquely defined by the points x1,...,xn
(I should have written the conditions x1=0 and x_n=t to that definition.)
Does this seem right?

I see. Thanks, martin!
I know how to understand the original question now

Does anyone know the tex command to make a tilde underneath a letter rather than above it?

\underset?

\underset{\tilde}{x} doesn't seem to work.

7:54 PM
http://latex.codecogs.com/gif.latex?\underset{\tilde}{u}

Should I use codecogs in an SE question?
(Does it even work?)

Or perhaps this: \underset{\widetilde{\hphantom{u}}}{u}

I'll change the notation... oh well

No, I was just showing how I got the image above.

@MartinSleziak: Aces! Thanks!
@robjohn: Thanks, too!

7:57 PM
It should look like this.

@MartinSleziak That's nicer, it has a bigger tilde

It does. I wonder why it has to be so complicated.

But, well, not everything that works in LaTeX (in codegogs) will necessarily work with mathjax.

The underset naturally makes things smaller.

@robjohn Could you perhaps be so kind and vote here math.stackexchange.com/questions/86093 to close as an exact duplicate?
(It's a long story - as you will notice in the comments.)

7:59 PM
@MartinSleziak I've entered my jury vote.

Thanks!

math.stackexchange.com/q/86134/1281 Didier: "yes, like a toddler anxiously clutching its parent's hand". Sometimes I find him like a child.

@Tim It does seem like a homework question that the OP doesn't want to tag that way.
I think that is Didier's reasoning for that comment.

Oh, @robjohn. I didn't know that. I meant that he is like a playful child sometimes.

I got that, but I was defending him in at least this case. :-)
I think I have spent too long trying to answer a question. However, I really think that the notation for d/dz and its "conjugate" are misleading.
Off to the store to get some groceries for the week. bbl

8:11 PM
This question makes very little sense. As far as I know A^2 is defined to be Spec k[x,y]...

8:24 PM
Hi everyone!

8:41 PM
@AsafKaragila: is this question okay as is?
or do I need to add information?

hello

Who is starring random stuff in here?
3
You, @AsafKaragila?

I'm back. Apparently one cannot leave the site alone for a few hours without drama erupting all over.
4

@HenningMakholm: What happened?

Um, everyone seem to be at each others' throats over the \binom question.

8:48 PM
@Matt Actually you can help by voting for "close as an exact duplicate" here: math.stackexchange.com/questions/86093
@HenningMakholm I don't think it's that bad...

@MartinSleziak: I would but I think I have to have a rep of more than 3000.

I thought you have that much.
Oh i see. I saw 3.9k here: chat.stackexchange.com/rooms/info/36/mathematics But that's probably cumulative from all sites.

9:22 PM
"For example in 𝔸^2_ℚ, the maximal ideal M=⟨x^2−2,y⟩⊂k[x,y] yields a closed point which is not rational (according to a theorem proved 25 centuries ago, albeit not in the language of scheme theory). "
Whoo!

Is it just me / my ISP, or is the site extremely sluggish at the moment?

@HenningMakholm I would not say extremely, but I noticed slowdown.

@HenningMakholm I just got back, but I will check the site.
@HenningMakholm MSE seems to be normal for me. What kind of slowness are you seeing?

9:37 PM
@HenningMakholm I am definitely not seeing that

Phew, I hope I am done with that answer... =)

@Srivatsan Got it done before it gets closed :-)

@robjohn - are you talking about the binomial coefficients one? I meant the little-O big-o answer... =)

It appears my DNS setup is on the fritz. Can't load anything else either.

@Srivatsan Ah, I was just looking at the most recent on your answers.

9:41 PM
@Matt Same here. I don't face any issues.
@robjohn My "on a serious note" note is longer than my answer ;)

@Srivatsan I have to choose "activity" and not "newest"

@Srivatsan Damn, I am so proud of myself that I'm blushing... =)
2

No comment. : )

Did you guys vote to close? Who wants to play the executioner?

@Srivatsan I'm on the jury, so I can't.

9:47 PM
@Matt, are you like too young to vote?

@Srivatsan: Yes (blush)

By young, I mean, less than the required reps

@Srivatsan How did you link back to your own comment? I don't seem to be able to do so.
I don't get an arrow at the right side of my own comments.

@robjohn Unimaginative solution. 1. Get the permalink, 2. Copy the message number, the number following the #, 3. Type : and then paste the number, followed by your actual comment.
4

Ah, so that is all the arrow does.

9:50 PM
@Srivatsan Boy, I've seen you in a while! Too busy?
would be an example.

@robjohn Like this?

@robjohn Don't get you. What's the magic in replying to me? :=)

@Srivatsan XD

@Srivatsan This message points to itself.
5

@Srivatsan clicko... when I clicked on the point to get the menu for the permalink, I think the page scrolled so I got your message instead.

9:52 PM
@Srivatsan, I don't get pinged when you @mention me in a comment to your own answer.

@HenningMakholm only one ping per message.

@HenningMakholm Oh, I didn't know that. Why so?
@robjohn No, that's only in comments. Here, I can ping @robjohn, @Matt, and @Srivatsan -- all in the same comment.

Give me a ping, Vasili. One ping only, please.
@Srivatsan Oh, that's good. I will definitely use that :-)

@Srivatsan Beats me -- I noticed @Henning in a comment to your big-O answer only by accident. Didn't get a ping.

@robjohn @Henning, you don't get pinged on this?

9:57 PM
@HenningMakholm Ok. Do you think, JM will be pinged by this: meta.math.stackexchange.com/questions/3257?

@Srivatsan @Henning, does this ping you?

@robjohn It does ping me. But the real test is Henning because my name comes ahead of his.

@robjohn @Srivatsan Now?

@Gigili It does.

@Srivatsan Yeah, it may be that on a chat message which starts with ':' only the first person gets pinged.

9:59 PM
@robjohn Yes.

@Srivatsan, @HenningMakholm, you probably both get this ping.