Calculating Sums of Reciprocal Squares and Fourth Powers

[dextercioby]

Compute the following:
$$\sum_{n=1}^{+\infty} \frac{1}{n^{2}} =...??$$
$$\sum_{n=1}^{+\infty} \frac{1}{n^{4}} =...??$$
.LINKS TO WEBPAGES WITH SOLUTIONS ARE NOT ALLOWED!

Daniel.

 

[polyb]

How about:

$$\zeta{(2)}$$

&

$$\zeta{(4)}$$

...or did you want a proof? or maybe a some numbers?

 

[The Bob]

$$\frac{1}{n^{\infty}}$$

This is a logical guess (well for me) so is likely to be wrong.

The Bob (2004 ©)

 

[T@P]

pi^2/6 and pi^4/90 respectively

 

[dextercioby]

I trust that any of could have taken a book on tables of series and products and find those values for $\zeta(2)$ and $\zeta(4)$.
The reason i thought of it as a (mathematical part of the) brain teaser is that i was waiting for a proof.
HINT:The roots are found in the same books where u found
$$\frac{\pi^{2}}{6}$$
and
$$\frac{\pi^{4}}{90}$$
,respectively.

Daniel.

PS.I believe i said "compute".

 

[T@P]

why doesn't someone just use a TI 89?

 

[dextercioby]

why doesn't someone just use a TI 89?

What's that?? It reminds me of the "Terminator" series in which the series for robots kept popping up from time to time...
Is it a calculator.??Computer software??Can it give a mathematical rigurous proof for those results?

Daniel.

 

[polyb]

As you well know the Riemann zeta function is some fun stuff and it can keep one very busy. So...

How about this dexter:

Let $$\zeta{(n)}$$ be defined as follows

$$\zeta{(n)}= \frac{2^n^-^1 \vert B_n\vert \pi^n}<br /> {n!}$$ where $$n\geq 2$$ and $$B_n = \frac{n!}{2\pi i} \oint \frac{z}{e^z-1} \frac{dz}{z^n^+^1}$$ (Bernoulli's Formula for his numbers)

for n= 2,4 then,

$$B_n = \frac{1}{6}$$ , $$\frac{-1}{30}$$

and plugging this into the former yeilds

$$\zeta{(n)}= \frac{\pi^2}{6} , \frac{\pi^4}{90}$$ respectively! Of course I also plugged your original equations into mathematica and got the same results!

 

[dextercioby]

Nicely done!Bravo!I'll wait a couple of days,maybe someone comes up with simpler solutions like the ones i have.If no,then i'll post my solutions,which are TERRIBLY SIMPLE!Anyway,they don't involve any complex functions integrations or residue formulas.

Daniel.

PS.Can you sove that integral,or did u pick it up from a book...?? I didn't find it in Abramowitz&Stegun.And neither the first formula.I can't look'em up in Gradsteyn&Rythzik,as the univ.'s library is closed. Can u prove that the Bernoulli numbers have the values they have?
In other words,can u come up with a consistent approach??
And one more question:Why is it called 'Bernoulli's formula for residues',if (i suppose Jakob,or maybe Johann) Bernoulli worked 100 before Augustin Cauchy introduced the term "residue" and gave his famous theorem??
And another one:you didn't specify the contour of integration,which is essential when integrating complex functions.

 

[TenaliRaman]

I aint got much time to write out my proofs , i will simply outline it.

1> If one divides the interval [0 degrees,90 degrees] into n equal intervals, then show that sum of the squares of tan of the angle at the center of each of the intervals is 2n^2-n. (The proof of this is fairly straightforward. Hint : use either euler's rule or De-moivre)
2> Once you have shown this, try to show that $$\sum \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}$$
3> Solving for zeta(2) is fairly simple. Once u have done this, there is fairly simple extension to finding zeta(4) and infact zeta(2n) in general (tho it involves a bit of messy algebraic simplifications).

In this very forum, i had given another fairly simple proof for zeta(2) using euler's sine product. I think it should be accessible by search.

-- AI

 

[polyb]

Can you sove that integral,or did u pick it up from a book...??

Actualy I picked it up from mathworld. Yes I can solve it but it has been a while and I would need to re-search it.

I didn't find it in Abramowitz&Stegun.And neither the first formula.I can't look'em up in Gradsteyn&Rythzik,as the univ.'s library is closed.

I found all this at mathworld under the Riemann zeta function, which has a really nice write up on it BTW. Worth the review. You should be able to find all of this in Abramowitz, though as you well know you have to be familiar with that book in order to use it properly.

Can u prove that the Bernoulli numbers have the values they have?
In other words,can u come up with a consistent approach??

Yes! I just need some time.

And one more question:Why is it called 'Bernoulli's formula for residues',if (i suppose Jakob,or maybe Johann) Bernoulli worked 100 before Augustin Cauchy introduced the term "residue" and gave his famous theorem??

MY BAD! I saw a complex contour intergral and my brain immediately jumped to residue theorum.

And another one:you didn't specify the contour of integration,which is essential when integrating complex functions.

It should be counterclockwise, sorry about that.

I will try the raw 'brute-force' appraoch and get back to you. If it is that simple then I should be able to find it.

 

[dextercioby]

I still find the complex function approach using Bernoulli numbers and Riemann zeta function rigurous and elegant at the same time.However,let us think for a little while to the ones that are not too familiar with complex integration and special functions.
The Euler's product for "sine" will give a marvelous proof that
$$\sum_{n=1}^{+\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$$
What about the other sum??Can someone come up with a simple elegant proof that it's got the result already stated??

Daniel.

PS.Beside the link at 'wolfram' site,i would indicate just for the heck of it this one:
http://www.gutenberg.org/dirs/etext01/brnll10.txt
Hope u guys have a good internet connection (like me ).

 

[T@P]

oh cmon a TI 89 is just a graphong calculator, put in the sum and it gives you the answer :)

 

[TenaliRaman]

dextercioby,
i know u don't want external link.
But i guess u wouldn't mind this one, its my own post at another forum, took me a long time to find it,
http://www.nrich.maths.org/discus/messages/8577/7180.html

-- AI(Arun Iyer)

 

[dextercioby]

There was not problem with the link whatsoever...

My first approach:

Use two expansions of the function
$$\frac{\sin x}{x}$$
around zero,one being the Taylor series:

$$\frac{\sin x}{x}=1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-\frac{x^{6}}{7!} +...$$ (1)
,and the other the famous Euler's sine product:
$$\frac{\sin x}{x}=(1-\frac{x^{2}}{\pi^{2}})(1-\frac{x^{2}}{4\pi^{2}})(1-\frac{x^{2}}{9\pi^{2}})...$$ (2)

Expand (2) and write it:

$$\frac{\sin x}{x}=1-\frac{x^{2}}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}+\frac{x^{4}}{\pi^{4}}\sum_{n,k=1;n\neq k}^{\infty} \frac{1}{n^{2}k^{2}}-\frac{x^{6}}{\pi^{6}}\sum_{n,k,l=1;n\neq k\neq l}^{\infty} \frac{1}{n^{2}k^{2}l^{2}}+...$$ (3)

Equate (1) and (3) to find:

$$\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$$ (4)

$$\sum_{n,k=1;n\neq k}^{\infty} \frac{1}{n^{2}k^{2}}=\frac{\pi^{4}}{120}$$ (5)

Square relation (4) to find:

$$(\sum_{n=1}^{\infty}\frac{1}{n^{2}})^{2}=\sum_{n=1}^{\infty}\frac{1}{n^{4}}+2\sum_{n,k=1;n\neq k}^{\infty} \frac{1}{n^{2}k^{2}}=\frac{\pi^{4}}{36}$$ (6)

From (5) & (6) u find simply that:

$$\sum_{n=1}^{\infty} \frac{1}{n^{4}}=\frac{\pi^{4}}{36}-2\frac{\pi^{4}}{120}=\frac{\pi^{4}}{90}$$

I don't personly like this method,coz i find working with sums rather difficult and unintuitive.My second method is more eleborate (requires some simple integrations),but i find more intuitive.


Daniel.