Deriving result from infinite series

[daring sequoia]

Hey y'all, I'm not too sure how I should go about this question. I tried multiplying both sides of the series eqn below but I can't think of how I could simplify it to get the above series

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[ashen hare]

try writing out first 4-5 terms of both sums

see if you spot a pattern

actually no wait its the other way around, give me a second

oh yeah, you can solve it from an equation

like if the first sum is equal to S and the second sum is equal to P, how does P + S/2 look like

ping me if you need further hints

[bleak hull]

Isn’t it moreso P-S? Doing this, something happens to the odd terms and so we can indeed solve an equation

[ashen hare]

well i mean

both ways lead to the same result

[bleak hull]

Huh i didn’t check you are most likely right

[daring sequoia]

ty! i tried writing writing out a few terms of P + S/2 though but idk if im doing this wrong but the series look more confusing now

when I wrote out P - S i can see that's equal to twice the sum of the even terms of P though

[fallow igloo]

@daring sequoia has given 1 rep to @ashen hare

[ashen hare]

wait i think i made an error

oh yeah i got it wrong, sorry

[bleak hull]

Yep

[coral harbor]

You could actually use fourier series to solve this

[ashen hare]

why

[bleak hull]

to solve the basel problem you can use parseval's theorem yes but thats not what is asked here

[worldly anchor]

The series is known as the Basel problem, and its solution is pi^2/6. The alternating series provided is related to it by considering the nature of alternating terms.

[
\sum_{j=1}^{\infty} \frac{1}{j^2} - \sum_{j=1}^{\infty} \frac{1}{(2j)^2}
]

[worldly walrus]

asura Ψ

[worldly anchor]

This expression represents the alternating series by including only odd terms in the sum for (\frac{1}{j^2}).