a number logic probem

Soroban

MHF Hall of Honor
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Hello, soyeahiknow!

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: .\(\displaystyle X \;=\;\dfrac{1}{2^2} + \dfrac{1}{4^2} + \dfrac{1}{6^2} + \hdots \)

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

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

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




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

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

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

 
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Oct 2009
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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
 

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MHF Hall of Honor
Mar 2010
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MHF Hall of Honor
Mar 2010
2,340
821
Chicago
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 2010
23
1
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
 

undefined

MHF Hall of Honor
Mar 2010
2,340
821
Chicago
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.
 
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