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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? .\(\displaystyle 2, 3, 5, 8, 13, 21,\:\hdots\)

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

Prime Zeta Function -- from Wolfram MathWorldI 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

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.

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.

Tonio

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.

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

For this problem, think about what happens when you add this series to the series that Soroban solved.

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