Mathematics

@TheGame Since I don't eat them I don't have one :P Though, I have one right now in the fridge, someone gave it to me, tis called Dairy Milk Nuts and Fruits.

@Alizter Me too :D ... though I end up eating and forgetting about math entirely.

Ice Boy

it depends on what you're reading

The Game

@Sawarnik I end up doing math and forgetting about eating entirely :/

Ice Boy

5:57 PM

me too^

Sawarnik

:// That's strange ://

Ice Boy

food for thought is stronger than food for the stomach

Chris's sis

@Alizter $$\displaystyle\sum_{n=1}^\infty\prod_{k=n}^m\frac1k=\frac{1}{m\cdot m!}$$

Alizter

@Chris'ssis change m to m+n?

on the top of the product

Chris's sis

@Alizter you said $m>n$

Alizter

6:00 PM

because I meant $\displaystyle\sum_{n=1}^\infty\prod_{k=n}^{m+n}\frac1k$

@Chris'ssis which makes the sums diverge

because 1 is empty rpoduct

you get the idea

if m is constant it will not always be $>n$

however m+n is what I meant

Chris's sis

@Alizter I'm not sure what you refer at.

$$\displaystyle\sum_{n=1}^\infty\prod_{k=n}^{m+n}\frac1k=\frac{1}{m\cdot m!},\space m\ge 1$$

Alizter

@Chris'ssis If $m$ is constant then there will be a time when $m<n$ therefore making pi the empty product which is 1 which will make the sum diverge

however the new formula is fine

The Game

@Chris'ssis From the pdf, A proof should be an epiphany from the Gods, not a coded message from the Pentagon :D

5

Chris's sis

@TheGame lol, yeah, sure. :-)

Ice Boy

It should, but rarely is.

Chris's sis

6:07 PM

Trying to find something nice I saw these days ...

Ice Boy

The Gods share very few of their epiphanies ...

... instead they encode them in conciseness.

Chris's sis

6

A: An uncanny inequality with Gamma function

Let $~g(x)=\dfrac{\Gamma'(x+1)}{\Gamma(x+1)}-\log x$. Now we have $$\eqalign{g'(x)&=-\frac{1}{x}+\sum_{k=1}^\infty\frac{1}{(x+k)^2}\cr &<-\frac{1}{x}+\sum_{k=1}^\infty\int_{k-1}^k\frac{dt}{(x+t)^2}\cr &=-\frac{1}{x}+\int_0^\infty\frac{dt}{(x+t)^2}=0 }$$ So $g$ is decreasing on $(0,+\infty)$. More...

@robjohn have you seen the question above?

The Game

@robjohn Do you know how to do the case $d_3$ ? math.stackexchange.com/a/951792/150347

Balarka Sen

6:26 PM

@TheGame Precisely

Chris's sis

hmmm, let me post a nice question (easy level) ...

Will Hunting

@BalarkaSen You should stop throwing tables

Ice Boy

why?

Balarka Sen

@WillHunting You should care less about trivias =P

Will Hunting

Throw something else instead, lol

Chris's sis

6:34 PM

Here is the foregoing one

Prove that

$$\sum_{n=1}^{\infty} \frac{ \log(n+1)}{n^2}= \zeta(2)\log\left( \frac{A^{12}}{2\pi e^{\gamma}} \right)+ \sum_{n=3}^{\infty} (-1)^{n+1} \frac{\zeta(n)}{n-2}$$

Balarka Sen

That's a cool identity, @Chris'ssis

@BalarkaSen Yeah.

@WillHunting Yup!

why?

@IceBoy Why?

6:43 PM

why not?

@WillHunting Yup!

@IceBoy Why not?

why?

OK, let me repost it with the proper requirement

Prove without pen and paper that

$$\sum_{n=1}^{\infty} \frac{ \log(n+1)}{n^2}= \zeta(2)\log\left( \frac{A^{12}}{2\pi e^{\gamma}} \right)+ \sum_{n=3}^{\infty} (-1)^{n+1} \frac{\zeta(n)}{n-2}$$

Sawarnik

@IceBoy Why not?

Ice Boy

6:46 PM

Yup!

Semiclassical

@Chris'ssis: What's $A$ in that?

Chris's sis

@Semiclassical Glaisher-Kinkelin constant

Sawarnik

@IceBoy Why?

Ice Boy

why not?

robjohn

@Chris'ssis Do you mean this one?

I believe so

Chris's sis

6:52 PM

@robjohn This one I mean math.stackexchange.com/questions/907472/…

robjohn

@Chris'ssis another obfuscation: putting $\log(1/e^\gamma)$ there...

@Chris'ssis yes, I believe I actually state that in an old answer.

Chris's sis

@robjohn I just grouped those things a bit :-)

@robjohn It's a nice question, really.

Alizter

@Chris'ssis I couldn't do it. Got too messy with pochhammers

I am exaughsted today though

trying to prove something about an elliptic equation

robjohn

@Chris'ssis That should not have been closed!

Semiclassical

elliptic as in pde or as in functions/integrals?

Chris's sis

6:55 PM

@robjohn Totally agree with you!

robjohn

I think that is the same as Gautschi or my inequality that I put in an answer yesterday

Chris's sis

@robjohn I voted for reopening the question.

John Jack

@Chris'ssis Okay thanks, only got response now.

Chris's sis

Who likes this question, might vote for its reopening.

8

A: An uncanny inequality with Gamma function

Let $~g(x)=\dfrac{\Gamma'(x+1)}{\Gamma(x+1)}-\log x$. Now we have $$\eqalign{g'(x)&=-\frac{1}{x}+\sum_{k=1}^\infty\frac{1}{(x+k)^2}\cr &<-\frac{1}{x}+\sum_{k=1}^\infty\int_{k-1}^k\frac{dt}{(x+t)^2}\cr &=-\frac{1}{x}+\int_0^\infty\frac{dt}{(x+t)^2}=0 }$$ So $g$ is decreasing on $(0,+\infty)$. More...

I mean it's a very good idea to reopen such a question, I'm sure there are still nice solutions to see.

The Game

I don't have enough rep to reopen :)

Chris's sis

7:05 PM

@TheGame No pb :-)

robjohn

I am getting a bit annoyed at the people who close good questions because there is no effort shown. I think that if a question is definitely not a homework question, it is perfectly valid for someone not to have a clue how to approach a problem they encounter in their studies. Asking for help should engender at least a suggestion or a hint, not a closure of their question as being off-topic. </rant>

9

Will Hunting

@ted If you were to rank English, French, German and Russian in the order of the amount of mathematical literature in these languages, would it be English > French > German > Russian, roughly speaking?

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$Mathematics$ Mathematics