An Infinite Sum of Reciprocals of Odd Squares

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[mega_mango]

pi²/6 - sum(1/(2i)²) = pi²/6 - 1/4*pi²/6 = 3/4*pi²/6 = pi²/8

[surenderkumar]

I have a challenge for all of you:
If 1/1^2+1/2^2+1/3^2+... infinity=pi^2/6 then evaluate:
1/2^3+1/6^3+1/12^3+...1/((n)(n+1))^3, as n tends to infinity.

[pieintegration]

I have a challenge:

Let's say there are 10 arbitrary positive real numbers a1, a2, a3, ..., a10 where each number is less than or equal to 1. Prove that the sum of the numbers has to be less than or equal to 9 plus the product of the 10 numbers.

[Blaqjaqshellaq]

Since pi is just under 22/7, pi^2 is just under 484/49, or a bit less than 9.9.

[MrGeorge]

But what can we do if we don't know about π²/6?

What about starting with a taylor series which can be manipulated to get this sum? I tried arctanh(x) with arctanh(x) = x + x^3/3 + x^5/5 + x^7/7 + ... But we need odd squares in the denominator so we have to find the antiderivative and have to adjust the function a little bit: arctanh(x)/x = 1 + x^2/3 + x^4/5 + x^6/7 + ...instead of just arctanh(x).

So if we find the antiderivative of arctanh(x)/x = x + x^3/3² + x^5/5² + x^7/7² + ... we will get the infinite sum for x = 1. At this point I got stuck...

[bkkboy-cmeb]

S₁=1/2²+1/4²+···+1/(2n)²+···
=1/2²(1+1/2²+···+1/n²+···)
=1/2²·π²/6=π²/24
S=(1+1/2²+1/3²+···+1/n²+···) - S₁
= π²/6 - π²/24
= π²/8

[dondunno]

-1/12 is real and you can't make it go away.
Or you could insist on the pure math perspective and say that every proton has infinite mass and every blackbody radiates infinite energy.
Is that the universe you live in?

[leonidfedyakov]

Ох, любите Вы растекаться мыслью по древу, как будто объясняете детям младшего школьного возраста, а не любителям математики )). Интересно было бы посмотреть Ваше решение базельской задачи, так как видео остальных «энтузиастов» на эту тему занудны в высшей степени.