# a number logic probem

• Jul 24th 2010, 08:55 PM
soyeahiknow
a number logic probem
It is a major theorem that

Attachment 18320

Thanks!
• Jul 25th 2010, 09:28 AM
Soroban
Hello, soyeahiknow!

Quote:

It is a major theorem that:

. . $\displaystyle \displaystyle{S \;=\; \sum^{\infty}_{n=1}\frac{1}{n^2} \;=\;\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \hdots \;=\;\frac{\pi^2}{6} }$

Use this to calculate: .$\displaystyle [1]\;\dfrac{1}{2^2} + \dfrac{1}{4^2} + \dfrac{1}{6^2} + \hdots$

We have: .$\displaystyle X \;=\;\dfrac{1}{2^2} + \dfrac{1}{4^2} + \dfrac{1}{6^2} + \hdots$

. . . . . . . .$\displaystyle X \;=\;\dfrac{1}{2^2}\underbrace{\bigg[\dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \hdots \bigg]}_{\text{This is }S}$

. . . . . . . .$\displaystyle X \;=\;\dfrac{1}{4}\cdot\dfrac{\pi^2}{6}$

. . . . . . . .$\displaystyle X \;=\;\dfrac{\pi^2}{24}$

Quote:

$\displaystyle [2]\;\dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{5^2} + \hdots$

Are the denominators primes? .$\displaystyle 2, 3, 5, 7, 11, 13,\:\hdots$

or a Fibonacci-type sequence? .$\displaystyle 2, 3, 5, 8, 13, 21,\:\hdots$

• Jul 25th 2010, 12:21 PM
soyeahiknow
I believe they are primes
• Jul 25th 2010, 12:34 PM
soyeahiknow
can u tell me why you took out the 1/(2^2) ?
• Jul 25th 2010, 01:04 PM
tonio
Quote:

Originally Posted by Soroban
Hello, soyeahiknow!

We have: .$\displaystyle X \;=\;\dfrac{1}{2^2} + \dfrac{1}{4^2} + \dfrac{1}{6^2} + \hdots$

. . . . . . . .$\displaystyle X \;=\;\dfrac{1}{2^2}\underbrace{\bigg[\dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \hdots \bigg]}_{\text{This is }S}$

. . . . . . . .$\displaystyle X \;=\;\dfrac{1}{4}\cdot\dfrac{\pi^2}{6}$

. . . . . . . .$\displaystyle X \;=\;\dfrac{\pi^2}{24}$

Are the denominators primes? .$\displaystyle 2, 3, 5, 7, 11, 13,\:\hdots$

or a Fibonacci-type sequence? .$\displaystyle 2, 3, 5, 8, 13, 21,\:\hdots$

I think the second one should be the sum of the squared inverses of the odd natural numbers, otherwise it seems to be pretty hard to calculate the sum over the primes...

Tonio
• Jul 25th 2010, 01:15 PM
undefined
Quote:

Originally Posted by tonio
I think the second one should be the sum of the squared inverses of the odd natural numbers, otherwise it seems to be pretty hard to calculate the sum over the primes...

Tonio

Prime Zeta Function -- from Wolfram MathWorld

id:A085548 - OEIS Search Results

EDIT: I should have added some comment to be clear; it's likely that tonio is right that the problem is misstated; nevertheless, if it is correctly stated then the above links are relevant.
• Jul 25th 2010, 01:20 PM
tonio
Quote:

Originally Posted by undefined

Thanx, I knew that, but I highly doubt an obviously beginner student can come up with a research about the prime zeta function...

Tonio
• Jul 25th 2010, 01:21 PM
undefined
Quote:

Originally Posted by tonio
Thanx, I knew that, but I highly doubt an obviously beginner student can come up with a research about the prime zeta function...

Tonio

Yes, I edited my post above with a comment, but not quick enough.
• Jul 25th 2010, 01:22 PM
soyeahiknow
wait wait,

i think there may have been a mistake.

The 2nd problem is just simply:

1/(1^2) + 1(3^2) + 1/(5^2)

So it is simply the odd numbers. So would I just do 1/(3^2) * S= 1/9 * (Pi^2)/6 = (Pi^2)/ 54? Thanks
• Jul 25th 2010, 01:26 PM
undefined
Quote:

Originally Posted by soyeahiknow
wait wait,

i think there may have been a mistake.

The 2nd problem is just simply:

1/(1^2) + 1(3^2) + 1/(5^2)

So it is simply the odd numbers. So would I just do 1/(3^2) * S= 1/9 * (Pi^2)/6 = (Pi^2)/ 54? Thanks

In post #2, Soroban factored out (1/2)^2 because it evenly divides each term, and because doing so allows us to express the sum as a product of already known quantities.

For this problem, think about what happens when you add this series to the series that Soroban solved.
• Jul 25th 2010, 01:33 PM
soyeahiknow
if I add this series to the one Soroban solves, it would just be back to the original series right?

OOOOO so all I have to do is ( original series - series with all the even numbers)= series for problem 2 ( the odd numbers). So (pi^2)/6-(pi^2)/24= (pi^2)/8

and what kind of problems are these called? Thanks so much and sorry for the typo in the beginning! I posted it at like 3 am.
• Jul 25th 2010, 01:36 PM
tonio
Quote:

Originally Posted by soyeahiknow
if I add this series to the one Soroban solves, it would just be back to the original series right?

OOOOO so all I have to do is ( original series - series with all the even numbers)= series for problem 2 ( the odd numbers). So (pi^2)/6-(pi^2)/24= (pi^2)/8

and what kind of problems are these called? Thanks so much and sorry for the typo in the beginning! I posted it at like 3 am.

You're right about the sum of the squared inverses of the odd naturals, and these series could be classified as ones related to the Riemann zeta function, a truly fascinating and intriguing series.

Tonio