# a number logic probem

#### soyeahiknow

It is a major theorem that

Thanks!

#### Soroban

MHF Hall of Honor
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? .$$\displaystyle 2, 3, 5, 8, 13, 21,\:\hdots$$

soyeahiknow

#### soyeahiknow

I believe they are primes

#### soyeahiknow

can u tell me why you took out the 1/(2^2) ?

#### tonio

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

#### undefined

MHF Hall of Honor

#### 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

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#### undefined

MHF Hall of Honor
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.

#### 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

#### undefined

MHF Hall of Honor
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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