12:08 AM
@PeterTamaroff Perhaps.
I prefer that it resolves by itself (and exercise), otherwise it takes quite some weeks to recover, and I had it before.
@PeterTamaroff Usually there is no real 'cause' in the sense that the things that cause it are actually quote normal, like bend to tie your shoes -like with me-. In that aspect it is caused by physical trauma, but the susceptibility is a whole different story. There does not seem to be a relationship between hard physical labour, but there is one with 'sitting professions', but confounding is lurking there...(...)
It can be quite reasonable to expect people with such professions to not exercise a lot and have thus worse condition of the muscles in their back causing larger forces on the vertebrae. Or that people that pick such professions are more likely to be less of physique anyway!

@JonasTeuwen I see. I do hope you get well soon, man.

Oh, it is not a biggie.
Just feels like some giant animal is gnawing on my legs sometimes 8-)).

@JonasTeuwen You will cycle a lot when you get to canberra. Trust me.

I need a bike that fits me 8-).
But I love to cycle.

12:15 AM
@JonasTeuwen there are plenty that will.

Good! Here they have to be ordered (and the Dutch are kinda the tallest bikers in the world).

@JonasTeuwen

But I suppose it will be available, just have to look for a bit.
I bet I cannot take my hiking shoes without customs tossing them in the bin.

@JonasTeuwen that is true don't play the fool with the customs here

Nice!
But they are like 500€'s. So I better leave them here eh.

12:17 AM
@JonasTeuwen also don't bring marijuana here

Yeah, like duh.

don't use a suitcase that you have stored marijuana in before.
the dogs will smell it.

But will find nothing.

yes but still
the customs will suspect you.

No, the reason alone that I might land in Singapore to transit is reason enough!
Sure, but that are the kind of mistakes I do not make. Just do what they say and it is over soon 8-).

12:18 AM
yes
and certain kinds of food you can bring in.
just declare.

And reply no to "Did you import 1) A bomb 2) Drugs 3) A corpse".

Like coffee is permissible.

I bring no foods.
Other than perhaps a candy bar from some airport.

@JonasTeuwen at least bring a speciality from holland.
Like I had a friend from mexico
he brought in so much tequila

Yea, but that would be marijuana.
Space cake.

12:19 AM
@JonasTeuwen no like some kind of food.

Space cake is food.
I can bring some Dutch liquor, but it is not so 'nice', but perhaps interesting.

@JonasTeuwen no space cake

@BenjaLim Sure.

@JonasTeuwen

@BenjaLim I would not even get out of this country you know... The drugs policies are not that much 'softer' than yours, they only allow for 'own use' and then only softdrugs.
And it is even illegal, they just do not prosecute.

12:22 AM
@JonasTeuwen you know a lot of people would rather go to Bali,Indonesia for schoolies than the gold coast
because even though in indonesia it's death for drugs

does Singapore enforce the death penalty for "intent to distribute" amounts of marijuana, or is it only narcotics?

you can easily find a lot of drugs on the main street in kuta
@Bitrex I don't know.
@Bitrex but they are very strict.

12:34 AM
@peoplepower

@PeterTamaroff yellow

@peoplepower I'm trying to solve this, formally.
I've gotten to $$|f_n(x)-f_m(x)|\leq \int_0^x |f_{m-1}-f_{n-1}|$$
Each $f_n$ is continuous for $n>1$
Since the space of continuous functions over a compact interval $[a,b]$ is complete wrt to the uniform norm, I intend to prove that the sequence is Cauchy, to prove the existence of the limit.
Then what I wrote informally will follow.

12:54 AM
@PeterTamaroff I am not seeing how this will address the point made in did's last comment.

@peoplepower I don't really care about that comment. Did I link to that?

Your informal thoughts will not follow, in particular "This suggests that, if $f_n(x) \to f(x)$ and $f_n'(x) \to f'(x)$..."
Unless you show that $f'(x)$ exists and is actually the limit of $f_n'$.

@peoplepower I intend to prove uniform continuity.

That is kind of automatic right? Your function is absolutely continuous.

The $f_n$ are continuous for $n>0$ differentiable if $n>1$
@JonasTeuwen Well, that is all we need, really.
...since the space of continuous functions over a compact interval $[a,b]$ is complete wrt to the uniform norm

12:59 AM
Okay, done. Good job.

@JonasTeuwen Why are they absolutely continuous?
What does absolutely continuous mean?

Wasn't it something about $\sum |f(x_k)-f(x_{k+1})|<\epsilon$ when $\sum \Delta x_k <\delta$?

And below that some properties of it.
Yes, but that is one ugly sucker of a definition if you ask me.
3

@PeterTamaroff Sorry, I was out running errands. I don't know who first came up with that proof, or even who first published it.

1:05 AM
@JonasTeuwen But why does uniform continuity follow automatically?

@Ethan: if you want a proof of the asymptotic behavior of the sum of slowly growing functions, repost the question.
@PeterTamaroff Absolutely continuous means that $\sum|f(x_i)-f(x_{-1})|$ is controlled by $\sum|x_i-x_{-1}|$

@PeterTamaroff Because it is like uniform continuity but 'many delta'.

@JonasTeuwen Hahah, but I'm talking about uniform convergence.

@PeterTamaroff Uniform continuity only requires that one delta $f$ be controlled by one delta $x$

@robjohn Yes, yes, I was just not following you because I thought I had written convergence, not continuity.

1:11 AM
@PeterTamaroff Oh

Also, uniform convergence will be easier as you do not allow much growth and have differentiability.

To sum up

$f_0$ is bounded away from zero.
$f_n$ thus is also bounded away from zero, except at $x=0$
$f_n$ is continuous for $n=1,2,3,\dots$
$f_n$ is differentiable for $n=2,3,4,\dots$

But, you can guess the limit right :-).
Just prove this is the limit.
You would use such arguments if it were too hard or impossible to find the limit.
(very general theorems)

$|f_n(x)-f_m(x)|<\int_0^x |f_{n-1}-f_{m-1}|$
@JonasTeuwen Right. I just think Didier's proof is very long winded.
I want a "nicer" proof.

By dominated convergence, your limit satisfies: $$f^2(x) = \int_0^x f(x)$$
Yes, but Didier is very... much paying attention to small details.

1:15 AM
@JonasTeuwen What is $f^n$ here?

His proof is very nice.

I have to take my son shopping. BBL

Didier very carefully finds good upper and lower bounds.

@JonasTeuwen Yes, I agree with that.
Looking at it twice.
It is pretty dope.
I guess I'm just not used to such proofs, that's all

I would have written it down differently. First give the argument, and then munge a bit on the details.
I go to bed.
Good night!

1:19 AM
@JonasTeuwen Bye. Sleep well.

1 hour later…
user19161
2:46 AM
It took a minute for my avatar to appear after entering the room, strangely long.

user19161
It also took a quarter minute for me to be able to edit the above line after typing it.

user19161
@robjohn Have fun shopping!

user19161
@JonasTeuwen Have fun sleeping!

3:31 AM
@JasperLoy Back. Wow; it seems to have been quite quiet.

@robjohn Duuuuuuuuude.
I would like to guess your age but if I over estimate you'll feel bad, and if I underestimate you'll get cocky and stuff.

@PeterTamaroff Hey
@PeterTamaroff I would?

@robjohn Of course. That is human behaviour 101.

@PeterTamaroff That's a little high...

@robjohn High?

3:38 AM
@PeterTamaroff 101

@robjohn I didn't tell the base though?
@robjohn I meant those "VISUAL BASIC FOR DUMMIES" kind of books.

@PeterTamaroff base? I thought 101 was your guess ;-)

Aren't they also called 101 books?

@PeterTamaroff I've heard advanced beginning courses called 101, but not books

@robjohn Oh. That!

3:41 AM
@PeterTamaroff I was trying to make a joke on the 101, but failed.
I'm getting too old for this sort of thing.

Must be 100 then

@robjohn Oh, that narrows it down to a 50-65 frame.

I just want to say, I think the Dot product is awesome.

@RavenDreamer All hail $\langle- ,-\rangle$!
It is asian, did you know?
Mongol, more precisely.

That'd probably make more sense if I had the mathscript thing installed.

3:46 AM
@RavenDreamer Shame on you.
@RavenDreamer Why do you say that?
@robjohn What was your mail again?
I want to even things out.

@PeterTamaroff "I only want to call this function if a certain vector has a positive Y component."
"Hm. But how to... can I just... a dot product here and..."
"Hey, it worked!"
"The Dot Product is awesome!"

@RavenDreamer I lost you right there!

Seriously? If you're not joking, I can try to explain it if you want.

@robjohn Did you see it?
@RavenDreamer I'm honest.

Okay. I am a programmer. I have two polygons, and I want to do something (call a function) when polygon A intersects the top of polygon B, but not the side.
"call this function" essentially means, "do something".

3:51 AM
@RavenDreamer Oh. When you say function I think $f:A\to B$

Well, same idea.
Take some input, produce some output.
By the way, I like your gravatar.

@RavenDreamer It is C&H awesomeness.
@robjohn Oh, you've forsaken me again.
@RavenDreamer So how does this polygon thing works?

@PeterTamaroff It's collision detection, using the Minkowski Sum B-A (depending on how much you like geometry, you may find this interesting )

@RavenDreamer This is a quintessetial TL;DR moment, but I'm scanning through.

Basically, given two convex polyhedra, I get a vector back that points in the direction of the minimum penetration depth, i.e., the minimum amount I'd have to move one of the shapes to not be colliding.

4:01 AM
@RavenDreamer Interesting.

Yeah.

@RavenDreamer Do you know anything about games?

A little.

I mean "modern" games.

user19161
@PeterTamaroff Well, he has mentioned it a number of times, you just did not pay attention.

4:06 AM
@PeterTamaroff Did you have a question about game development?
If nothing else, my cousin is an actual game developer, so I can forward the question :P

@RavenDreamer Not particularily. But I'd like to know what you think about game prices.

user19161
@PeterTamaroff I think games are boring.

@PeterTamaroff Game Prices I don't have much of an opinion on. $60 is a lot for a game, but if it's a decent game, I'm okay with paying it. @JasperLoy Then you don't understand games. What I hate, is DLC. 4:09 AM @RavenDreamer What is DLC? Down loadable content. user19161 @PeterTamaroff I am not Professor Nash. Basically, you can buy more game. @RavenDreamer Oh, OK. It's a two part issue for me. First, down loadable content is generally much less value / time spent. 4:10 AM @RavenDreamer What you think about piracy? I used to like games when I was younger, but as I got older I started to lose interest... I would have fallen all over Skyrim when I was like, 19 but now...eh I don't want to pay$10 for another 2 hours of game (esp. since the first $60 can usually get you 10-20 hours) Chrono Trigger was the Skyrim of my day And the 2nd, is the bull@#$* devs who use 'preorder' bonuses, or 'vendor specific' content (i.e., you only get this gun if you buy from gamestop)

user19161
@RavenDreamer Games you mean, game would mean something else.

4:11 AM
@PeterTamaroff I think always on DRM sucks. Gabe Newell (CEO of Valve) summed it up pretty well for me: "Piracy is a service problem".

@RavenDreamer In what sense is it a service problem? What do you mean?

And by "always on", I mean, "We require you to have a constant internet connection to play this game. If you lose connection for whatever reason, we will kick you out of the game"

For example, I would totally buy a license for Batman Arkham City if I could. I already have the friggin files, so I really just need the license.
But then on Amazon they want me to download the files too, which is like 8GB.

user19161
I have played almost no computer games in my life.

Does this make sense to you?
Some people seriously object to the statement that it is forbidden to claim the truth of any theorem (that uses any form of arithmetic, no matter how simple).

I think this follows trivially from Gödel's incompleteness theorem. Claiming that such theorem would be true, amounts to creating a new theorem consisting of the original theorem and a statement about its consistency. This new theorem is automatically inconsistent:

"For any formal effectively generated theory T including basic arithmetical truths and also certain truths about formal provability, if T includes a statement of its own

4:15 AM
@RavenDreamer Oh. That really sucks.

@PeterTamaroff Most pirates pirate because they are unable or unwilling to acquire the game legally. Most media companies assume piracy = lost sales, but my opinion is that if you prevent piracy, you don't generate sales, because the choice is often. "pirate or don't play" and not "pirate or buy".

I hate it that I have to log in or connect to itnernet to play a game.

Yes.

user19161
@RavenDreamer In fact piracy can be argued to lead to higher profits, because they help advertise.

4:17 AM
@JasperLoy Exactly!
There are also studies for music piracy, at least, which say that the biggest pirates are also the biggest spenders.

I agree with you in some points but not all of them

@PeterTamaroff Thanks, haha. I just cloned Bit.Trip Beat and composed my own music to go with it.

user19161
@RavenDreamer I don't save PDF copies of books, but I do review them before deciding whether to buy.

If someone shows you how to get torents or stuff you might decide to stop buying dvds and pirate instead
The problem in my mind is that pirating stuff helps to spread piracy

4:20 AM
@Khromonkey The problem with pirating is that it is gray-legal, and you can run into malware, all kinds of nasty stuff.

oh come on

The challenge is for content providers to offer their content in a way that makes it more valuable than pirating

user19161
If I could not review those books before buying, I would never buy them.

I know a guy who has 1tb of pirated stuff on his computer and it works fine

user19161
4:21 AM
And even in a university library there are many important books missing.

getting a tv show on kat.ph is easier than going to the store and buying the season ive been told

user19161
@Khromonkey I have none on my computer. =)

@Khromonkey Right. So the solution is to encourage buying the season, instead of getting the pirated version.

why not both?

user19161
Did any of you here watch the entire Wonder Years series?

4:23 AM
Because preventing piracy outright is an impossible goal.

what?
You can do both right?

(And also because the MPAA / RIAA don't provide viable alternatives, esp. to foreign audiences)

you can both get the pirate version and get other people to buy it

@RavenDreamer Do you agree that playing a game for the graphics is like watching... ?

@Khromonkey The issue is that it becomes (effectively), an arms race between the pirates and the people trying to prevent piracy, and this arms race inevitably leads to collateral damage against paying customers (i.e., the "Always Online" DRM I was talking about earlier) that the pirated versions don't have, thus making the pirated version even better in the consumer's mind (easier, don't have to worry about losing connection, etc).
2
@PeterTamaroff like watching what?

4:26 AM
@RavenDreamer Google "playing a game for the graphics is like watching"
I'm trying to keep things PG13

I dont agree

Um. I'm not sure how I feel about that.

It depends on the genre

But then again, I'm not a graphics buff.
I'd say the inverse is true.

For example, in shooters or car games I think the graphics are really important

4:27 AM
"Not playing a game because of the graphics is like not watching..."

so then seeing the ----- for the ----- is like playing a game for the graphics

My criteria is simple: Are you enjoying it? If y, then continue. Else, stop playing.

In my opinion graphics can not make a great game
however they can make a terrible game playable

I can respect that.

I mean, I sometimes enjoy playing racing games because the cars look awesome
Although the game is nothing special really

4:32 AM
Does anyone know the game Soulbringer?
I think it kicks ass.

My favorite game is final fantasy xii
I played it when I was 12
final fantasy xiii is a pile of junk for uncoordinated retarded toddlers

I played Final Fantasy 7 in college :B
How does one integrate something like: $u(x) = \int \frac{f(y)}{|x - y|}d^3y$
I'm just a dumb EE, and that's throwing me for a loop.

@Bitrex The notation is a little strange there.

It's supposed to be how you find a solution to the 2D Laplacian given the fundamental solution
Green's function and stuff
sorry, 3D laplacian
x and y must be vectors
I think
I dunno
Oh, it's just like solving the equation for electric potential in 3 dimensions, it's easy to do in spherical coordinates

5:08 AM
graphics tend to obscure the truth. im biased since i used to develop MUDs
in txt games what is is simply said :)

when I was a kid I tried to write a text adventure in BASIC, but I was lazy about crafting room descriptions
so I had it take place in a post-apocalyptic nuclear wasteland
"You are standing in a nuclear wasteland."
>w

What does OP stand for again?

"You are standing in a nuclear wasteland."

@JayeshBadwaik: Thank you very much for the email :)

@bitrex geez we have an uncommon amount in common
same age same genres same starting points also math is why we are both here i assume
the MUD i started is armageddon.org which is post-apocalyptic among other things
also my first programming project as a young child was a BASIC text adventure inspired by Zork

5:23 AM
Hah!

Not much there!
Sorry!
I remember reading a book on how to do Zork-like games in BASIC
I remember circa 1989 or going to a library a couple towns over that had the only copy in my area
or so

mine was called "dragon" i guess it was hobbit-cartoon-based i remember writing the basic code in my notebook during recess in 3rd grade

5:56 AM
i had those basic program books too 1986 or so... atari st

6:27 AM
in a hotel bar right now, gotta sign out, @bitrex will keep seein u around i here i hope

1 hour later…
7:46 AM
Hi all
Experimentally I've come across the fact that $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n(m^2+n^2)}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n^2(m^2+n^2)}$. I wonder how I can prove that ... (without computing the double sums separately)

8:22 AM
I'm not that inspired this morning.

that shouldn't be possible, by the comparison test
but wolframalpha says they are both the same

@anon W|A tells that both double sums tend to $pi^4/72$
@anon the result is astonishing to me, and I'd really like to know what actually happens there since both of them coverge to the same value. Initially I thought my input was wrong, but then I admitted the surprising real fact. :-)

Wolfram Alpha sometimes gives answers that are incorrect as Mathematica also does, but in Mathematica this is usually accompanied by error messages. This is what I got in Mathematica 8:
$\sum _{n=1}^{\infty } \left(\sum _{m=1}^{\infty } \frac{1}{n \left(m^2+n^2\right)}\right)$
becomes:
$\sum _{n=1}^{\infty } \frac{\pi n \coth (\pi n)-1}{2 n^3}$
=1.9887084003563...

$\sum _{n=1}^{\infty } \left(\sum _{m=1}^{\infty } \frac{1}{n^2 \left(m^2+n^2\right)}\right)$
tends to:
$\frac{\pi ^4}{72}$
=1.35290404213...

8:40 AM
I wonder what fallacious series manipulation alpha used.

@MatsGranvik: interesting. Actually, they are proving to be different.

Note that the original series CS posted had n indexed on the inside and m on the outside.

@anon right!

I still get 1.35642 (+0.i) versus 1.3529 when I use NSum on the correctly indexed series.

Switching them I get $\frac{\pi ^4}{72}$ in both cases.

8:45 AM
@anon: good observation.
So, they both take the same value.

According to Mathematica and Alpha..

@Chris'ssister:
Sum[Sum[1/(n^2*(m^2 + n^2)), {n, 1, Infinity}], {m, 1, Infinity}]
Sum[Sum[1/(n*(m^2 + n^2)), {n, 1, Infinity}], {m, 1, Infinity}]

Does anyone may try it with maple pls? :D (I have no computational software unfortunately).

A strange thing happens though when you look at the differences between their partial sums. If both (outer) series converge to the same limit, we should expect the difference between their partial sums to converge to 0 - but it doesn't.

@anon: right

8:48 AM
trying it with maple 13
maple was unable to do the first sum
nevermind, spacing issue

@anon: thanks.

holy moley you are dumb maple

:)))))

maple is conveniently forgetting what the 'evalf' command is supposed to do when it comes to the first series, but the second it gives the numerical value for pi^4/72

@anon: however, even if the indexing of series is changed the values should be the same because of the symmetry, isn't it?

8:56 AM
huh? interchanging limits requires uniform convergence or something like that, and I don't see what symmetry you are referring to (the summand is not preserved under interchange of n and m for instance)

user19161
Now you know why I don't use Maple, or Mathematica, or Matlab...

user19161
Anyway, they all cost Money.

At least Mathematica makes mathematical mistakes, instead of ludicrously opaque and perverse interpretations of commands, and spontaneously deciding not to obey commands.
looking at you, maple

user19161
So far, I only use Calculator. For function graphs, PSTricks in TeX.

@anon: $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n(m^2+n^2)}=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{n(m^2+n^2)}$ shoud be the same due to the symmetry

8:59 AM
what symmetry, and why?

@anon: take n=50 and m=50 in W|A and see that you get the same values.(or any other finite values easy to compute)

so?

user19161
Hello @mats nice Christmas hat there!

@JasperLoy thanks.

@anon: because both series are the same thing.

9:00 AM
"it"?

@anon: yeah. am I wrong?

There are doubly-indexed series that give different values depending on which index you run over first. If you only look at the "diagonal," the partial sums of partial sums, they will both look the same of course. It is the same as how, in general, the limits $$\lim_{x\to \infty}\lim_{y\to \infty}f(x,y),\quad \lim_{y\to\infty}\lim_{x\to\infty}f(x,y),\quad\lim_{x\to\infty}f(x,x)$$ can all be different.

OK. Let me recheck things.
@anon: agree with you 100%.

@JasperLoy, those CAS (and matlab) may cost money, but there are loads of libraries that are pretty good that don't...

As further proof, you see how Mats inputs gave different outputs when he used a different indexing than the original one you posted, whereas with the original indexing there are equal outputs.

9:08 AM
@anon: right.

@JasperLoy, in fact MuPad which is now part of matlab was freeware back in the day
that's what got me started with CAS

holy crap it is 3am
I thought it was like 12am

10am here

@anon: maybe I should post it on main to ask for more opinions.

9:15 AM
@PeterSheldrick MuPad was great when it was free ... seems a long time ago

yes :(
@OldJohn, nice that you are so eclectic in your CAS usage!

Now I find that Pari/gp suits all my number theory needs

mhm
that pops up loads of times on this site

eclectic sounds like me

big plus that it is open source

9:18 AM
not sure pari is great for much else - it was created by a number theorist, I think, so it is a bit biased - and free!

yeah it's specialized

I downloaded sympy the other day - but not had time to find out how to use it

as opposed to for example sympy which may do some basic number theory thing, or maybe not
@OldJohn, good!
it is probably mainly for python enthusiasts

I have avoided python - and ASP - I hate snakes :)

it is named after monty python - even though most graphics for python use the snake -.-

9:24 AM
@PeterSheldrick I never knew the Monty Python connection - great :)

@anon: do you admit this $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m(m+n)}=\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{n(m+n)}$ is correct (it's just an example)?

Chill dude: WHATSUP
Mathematician: The least upper bound of a set.

@Chris'ssister The inner sum $\sum_{n=1}^\infty \frac{1}{m(m+n)}$ isn't convergent for a fixed $m$, is it?

no, it isn't, though I think the outer sum on the right is divergent (while the inner sums are convergent)

I can't believe those 2 series in the question on main really have the same sum
@anon agreed

@Chris'ssister yes, that's the one I mean

@anon: or $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2(m^2+n^2)}=\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{n^2(m^2+n^2)}$. They must have the same limit due to the symmetry.

@OldJohn neither can I - comparison test on the summands shows that the inner sums on the left are always greater than those on the right, so that the outer sums must be inequal by a second application of the comparison test.

@anon absolutely - that's what I concluded

@Chris'ssister what symmetry? do you think every doubly-indexed series exhibits "symmetry" because the partial sums of partial sums are the same no matter which way you sum them (obviously because finite sums can be reindexed willy-nilly)?

9:38 AM
@anon: I didn't refer to every doubly-indexed series, but to my examples.

Even singly-indexed series can be tricky suckers when you rearrange them - unless absolutely convergent

Surely $$\sum_{n=1}^N\sum_{m=1}^Ma_{n,m}=\sum_{m=1}^M\sum_{n=1}^Na_{n,m}$$ for every $N,M$ no matter what the sequence $a_{n,m}$ looks like. However, it is not generally true that $$\lim_{N\to\infty}\sum_{n=1}^N\lim_{M\to\infty}\sum_{m=1}^Ma_{n,m}=\lim_{M\to\i‌​nfty}\sum_{m=1}^M\lim_{N\to \infty }\sum_{n=1}^Na_{n,m}.$$
@Chris'ssister And you cited "symmetry" as justification, but your only indication of what that means is that the partial sums of partial sums can be indexed either way and the value is the same, which is hardly particular to your examples.

@anon: well, $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2(m^2+n^2)}=\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{n^2(m^2+n^2)}=\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{1}{n^2(m^2+n^2)}$

Ragib has given a very neat evaluation of the RHS on main

Ragib says the same thing, but I think technically more needs to be said for it to be justified.

9:42 AM
I think Ragib is fine - all the series are absolutely v=convergent

Right.
I guess I'm just uppity because the behavior is so strange.

@anon: Ragib also used the symmetry.

which involved reindexing, possible due to absolute convergence
not always applicable

@anon: agree that it's not always applicable.

@anon Being uppity is a good thing for a mathematician - or a least a really healthy dose of scepticism :)