Question

show that \[\large\sum\limits_{a,b=1}^{\infty} \dfrac{1}{a^2b^2} = \dfrac{5}{2}\]

for relatively prime \(a,b\)

A not so useful hint can be found @
http://openstudy.com/users/ganeshie8#/updates/54e1ae2ee4b0b0ad88553f82

Answer 1

\(\large \begin{align} \color{black}{\sum\limits_{a,b=1}^{\infty} \dfrac{1}{a^2b^2} = \sum\limits_{a=1}^{\infty} \dfrac{1}{a^2} \times \sum\limits_{b=1}^{\infty} \dfrac{1}{b^2} \hspace{.33em}\\~\\}\end{align}\)


so this is equivalent to this ?

Answer 2

thats a good start!

Answer 3

ok so this identity is true alright .

Answer 4

\[\sum\limits_{a,b=1}^{\infty} \dfrac{1}{a^2b^2} = \sum\limits_{a=1}^{\infty} \dfrac{1}{a^2} \times \sum\limits_{b=1}^{\infty} \dfrac{1}{b^2} = \left(\sum\limits_{c=1}^{\infty} \dfrac{1}{c^2} \right)^2 \]

the variables are just dummy..

Answer 5

how can we write \(c\) in terms of \(a\) and \(b\)

Answer 6

\(\large \begin{align} \color{black}{\sum\limits_{a,b=1}^{3} \dfrac{1}{a^2b^2} = \sum\limits_{a=1}^{3} \dfrac{1}{a^2} \times \sum\limits_{b=1}^{3} \dfrac{1}{b^2}=\left(\sum\limits_{c=1}^{3} \dfrac{1}{c^2} \right)^2 \hspace{.33em}\\~\\}\end{align}\)

if i consider small example like this then how is \(c\) related to \(a\) and \(b\)

Answer 7

Haha I guess I would have to say it's pi^4/36 but apparently I need to look at it differently hmm

Answer 8

i think we need to use that result..
the original sum only involves the relatively prime pairs a,b

Answer 9

Ahhh right, I forgot about that.

Answer 10

@mathmath333 you seem to be on right track.. keep going :)

Answer 11

Hmm I don't know if this is a true statement anymore since a and b are relatively prime hmmm.

\[\ \sum\limits_{a=1}^{\infty} \dfrac{1}{a^2} \times \sum\limits_{b=1}^{\infty} \dfrac{1}{b^2} = \left(\sum\limits_{c=1}^{\infty} \dfrac{1}{c^2} \right)^2 \]

Answer 12

Exactly! i think we can start wid that general result and see if we can split the sum based on the given condition..

Answer 13

below is true if we run s,t over all positive integers \[\ \sum\limits_{s=1}^{\infty} \dfrac{1}{s^2} \times \sum\limits_{t=1}^{\infty} \dfrac{1}{t^2} = \left(\sum\limits_{s=1}^{\infty} \dfrac{1}{s^2} \right)^2 \]

Answer 14

i dont understand why u say and b are relatively prime i thought a and b are nothing but indeces

Answer 15

oh i see

Answer 16

yes, in the sum, consider only the indices pairs (a,b) that are relatively prime

Answer 17

it is just a generally accepted notation to put conditions on terms in sum..

Answer 18

i m still unable to understand how will sigma satisfy ur condition

Answer 19

of a and b coprime

Answer 20

\[\large \begin{align} \color{black}{\sum\limits_{a,b=1}^{3} \dfrac{1}{a^2b^2} = \sum\limits_{a=1}^{3} \dfrac{1}{a^2} \times \sum\limits_{b=1}^{3} \dfrac{1}{b^2}=\left(\sum\limits_{c=1}^{3} \dfrac{1}{c^2} \right)^2 \hspace{.33em}\\~\\}\end{align} \]

this is tre when there are no conditions on a,b @mathmath333

Answer 21

we need to somehow use this and extract the sum when a,b are relatively prime

Answer 22

yes that is true ,no conditions apply

Answer 23

\[\large \begin{align} \color{black}{\sum\limits_{a,b=1}^{3} \dfrac{1}{a^2b^2} =\left(\sum\limits_{c=1}^{3} \dfrac{1}{c^2} \right)^2 \hspace{.33em}\\~\\}\end{align} \]

lets evaluate this and see if we get same value on both sides

Answer 24

yes u know its true when condition aren;t there

Answer 25

yes

Answer 26

but when conditions do apply how will the summation process behave

Answer 27

One way to work it is by subtracting the terms related to a,b not relatively prime from above sum

Answer 28

something like this :

\[\sum\limits_{a,b=1}^{\infty} \dfrac{1}{a^2b^2} = \left(\sum\limits_{c=1}^{\infty} \dfrac{1}{c^2} \right)^2\]

So
\[\sum\limits_{\gcd(a,b)=1, a,b=1}^{\infty} \dfrac{1}{a^2b^2} = \left(\sum\limits_{c=1}^{\infty} \dfrac{1}{c^2} \right)^2 - \sum\limits_{\gcd(a,b)\ne1,~a,b=1}^{\infty} \dfrac{1}{a^2b^2} \]

Answer 29

i think this stuff is either above my reach

Answer 30

found this , might be helpful

http://www.wolframalpha.com/input/?i=%28%5Cfrac%7B%5Cpi%5E2%7D%7B6%7D%29%5E2%3D%5Cfrac%7B5%7D%7B2%7D%5Csum%5Climits_%7Bk%3D1%7D%5E%7B%5Cinfty%7D+%5Cdfrac%7B1%7D%7Bk%5E4%7D+

Answer 31

Thats it!

Answer 32

we just need to put all the pieces together..

Answer 33

which is tricky ofc

Answer 34

still trying to understand the process

Answer 35

i got idea from zeta function
ok will write down in morning

Answer 36

I'll post a solution by 5 PM IST today incase if this question still remains open :)

Answer 37

OMG CONEXUS BOTS

Answer 38

why are there so many people here

Answer 39

lol are they connexus teachers

Answer 40

they need to get laaaiiidddd XD

Answer 41

that comment made me laugh but il need to remove that haha

Answer 42

i think this will be having some telescooping series .

Answer 43

\(\large (\frac{\pi^2}{6})^2-x=\frac{5}{2}\)

Answer 44

wolfram gave some "digamma function" earlier, which i don't know

Answer 45

it seems from your wolfram link we have
\[\sum\limits_{a,b=1}^{\infty} \dfrac{1}{a^2b^2} = \left(\sum\limits_{c=1}^{\infty} \dfrac{1}{c^2} \right)^2 = \left(\frac{\pi^2}{6}\right)^2 = \dfrac{5}{2}\sum\limits_{k=1}^{\infty} \dfrac{1}{k^4} \]

?

Answer 46

yes

Answer 47

so we need to show that
\[\sum\limits_{a,b=1}^{\infty} \dfrac{1}{a^2b^2} ~~=~~ \sum\limits_{k=1}^{\infty} \dfrac{1}{k^4} \times\color{Red}{ \dfrac{5}{2}} ~~=~~ \sum\limits_{k=1}^{\infty} \dfrac{1}{k^4} \times \color{Red}{\sum_{\substack{a,b=1\\\gcd(a,b) =1}}^{\infty} \dfrac{1}{a^2b^2} }\]

Answer 48

\(\large \begin{align} \color{black}{\sum_{\substack{a,b=1\\gcd(a,b) =1}}^{3} \dfrac{1}{a^2b^2}\hspace{.33em}\\~\\}\end{align}\)


what will be the first 3 values for this

Answer 49

\[\large \begin{align} \color{black}{\sum_{\substack{a,b=1\\gcd(a,b) =1}}^{3} \dfrac{1}{a^2b^2}\hspace{.33em}\\~\\}\end{align}\]

is same as

\[\large \begin{align} \color{black}{\sum\limits_{a=1}^3\sum_{\substack{b=1\\gcd(a,b) =1}}^{3} \dfrac{1}{a^2b^2}\hspace{.33em}\\~\\}\end{align}\]

Answer 50

\[\begin{align}\sum\limits_{a=1}^3\sum_{\substack{b=1\\gcd(a,b) =1}}^{3} \dfrac{1}{a^2b^2}&= \left(\dfrac{1}{1^21^2} + \dfrac{1}{1^22^2} +\dfrac{1}{1^23^2} \right) \\~\\~\\
&+ \left(\dfrac{1}{2^21^2} +\dfrac{1}{2^23^2} \right) \\~\\~\\
&+ \left(\dfrac{1}{3^21^2} +\dfrac{1}{3^22^2} \right) \\~\\~\\

\end{align}\]

Answer 51

Notice without gcd(a,b)=1 conndition, we will get 9 terms in the sum
but with that condition there are two terms missing : \(\dfrac{1}{2^22^2}\) and \(\dfrac{1}{3^23^2}\) because \(\gcd(2,2) \ne 1\) and \(\gcd(3,3)\ne 1\)

Answer 52

Oh u revealed

Answer 53

I cant read fully latex don't appear on time but I might have same idea I thought of it like this for assuming gcd of a,b is not 1 then I would have this (sum 1/a^2)^2 which is the same as (zeta 2 )^2 with expanding some terms assuming ur series S I would got S= (zeta 2/ zeta 4 )^2= 2/5
Its annoying posting from phone :(
But looking forward to ur method seems the same :o

Answer 54

Lol hbd kai :p

Answer 55

And yet this
(Sum 1/ a^2 * sum 1/b^2 ) = (sum 1/b^2)^2
From kai comment but from gcd (a,b ) is not 1 its for a,b with all possibilities of gcd which is the same as (zeta 2 )^2

Answer 56

Just adding extra counter for all gcd hmmm idk if any one understand me lol but I don't have laptop xD to write neat

Answer 57

the solution i am having similar to that... could you post the detailed solution :)

Answer 58

mystery continous..

Answer 59

Wow very neat :o

Answer 60

It make sense to me and we don't need zeta xD

Answer 61

Like u only used it in final answer u didn't made mass expantion this question made my week lol

Answer 62

That is exactly same as your idea.. I just tried to make it as look as simple as possible by avoiding zeta. Below is true as you said earlier :
\[\sum_{\substack{a,b=1\\\gcd(a,b) =1}}^{\infty} \dfrac{1}{a^2b^2} = \dfrac{\zeta(2)^2}{\zeta(4)} = \dfrac{5}{2}\]

Answer 63

Fixed a typo :

We know below is true
\[ \left(\frac{\pi^2}{6}\right)^2 = \left(\sum\limits_{c=1}^{\infty} \dfrac{1}{c^2} \right)^2 =\sum\limits_{x,y=1}^{\infty} \dfrac{1}{x^2y^2} \tag{1}\]

Next notice that all the positive integer pairs can be split into classes based on gcd.
Like, all the pairs \((x,y)\) whose gcd is \(2\) exist in a class \(d_2\),
all the pairs \((x,y)\) whose gcd is \(3\) exist in a class \(d_3\),
etc...
each pair exists in one and only one gcd class. So we can split the sum in \((1)\) as below :
\[ \left(\frac{\pi^2}{6}\right)^2 = \left(\sum\limits_{c=1}^{\infty} \dfrac{1}{c^2} \right)^2 =\sum\limits_{x,y=1}^{\infty} \dfrac{1}{x^2y^2} = \sum\limits_{d=1}^{\infty}\sum_{\substack{x,y=1\\\gcd(x,y) =d}}^{\infty} \dfrac{1}{x^2y^2} \tag{2}\]

Since \(\gcd(x,y) = d\), we can write \(x,y\) as \(x = ad,~ y = bd\) for some relatively prime integers \(a,b\) and \((2)\) becomes :
\[\begin{align} \left(\frac{\pi^2}{6}\right)^2 &=\sum\limits_{d=1}^{\infty}\sum_{\substack{a,b=1\\\gcd(a,b) =1}}^{\infty} \dfrac{1}{(ad)^2(bd)^2} \\~\\
&=\sum\limits_{d=1}^{\infty} \sum_{\substack{a,b=1\\\gcd(a,b) =1}}^{\infty} \dfrac{1}{d^4} \dfrac{1}{a^2b^2} \\~\\
&=\left(\sum\limits_{d=1}^{\infty} \dfrac{1}{d^4} \right) \left(\sum_{\substack{a,b=1\\\gcd(a,b) =1}}^{\infty} \dfrac{1}{a^2b^2} \right)~~~\color{red}{\star }\\~\\
&= \left(\dfrac{\pi^4}{90} \right) \left(\sum_{\substack{a,b=1\\\gcd(a,b) =1}}^{\infty} \dfrac{1}{a^2b^2} \right)\\~\\
\end{align}\]

So we have
\[ \left(\frac{\pi^2}{6}\right)^2 = \left(\dfrac{\pi^4}{90} \right) \left(\sum_{\substack{a,b=1\\\gcd(a,b) =1}}^{\infty} \dfrac{1}{a^2b^2} \right)\]
yielding
\[\dfrac{5}{2} =\sum_{\substack{a,b=1\\\gcd(a,b) =1}}^{\infty} \dfrac{1}{a^2b^2} \]

\(\color{red}{\star}\) : refer to the hint in main question

Answer 64

cool