An Infinite Sum of Reciprocals of Odd Squares

Подписаться 151 тыс.

Просмотров 2,5 тыс.

50% 1

🤩 Hello everyone, I'm very excited to bring you a new channel (aplusbi)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
/ @sybermathshorts
/ @aplusbi
❤️ ❤️ ❤️ My Amazon Store: www.amazon.com...
When you purchase something from here, I will make a small percentage of commission that helps me continue making videos for you.
If you are preparing for Math Competitions and Math Olympiads, then this is the page for you!
CHECK IT OUT!!! ❤️ ❤️ ❤️
❤️ A Differential Equation | The Result Will Surprise You! • A Differential Equatio...
❤️ Crux Mathematicorum: cms.math.ca/pu...
❤️ A Problem From ARML-NYSML Math Contests: • A Problem From ARML-NY...
❤️ LOGARITHMIC/RADICAL EQUATION: • LOGARITHMIC/RADICAL EQ...
❤️ Finding cos36·cos72 | A Nice Trick: • Finding cos36·cos72 | ...
⭐ ⭐ Can We Find The Inverse of f(x) = x^x? • Can We Find The Invers...
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/...
Follow me → / sybermath
Subscribe → www.youtube.co...
⭐ Suggest → forms.gle/A5bG...
If you need to post a picture of your solution or idea:
in...
#radicals #radicalequations #algebra #calculus #differentialequations #polynomials #prealgebra #polynomialequations #numbertheory #diophantineequations #comparingnumbers #trigonometry #trigonometricequations #complexnumbers #math #mathcompetition #olympiad #matholympiad #mathematics #sybermath #aplusbi #shortsofsyber #iit #iitjee #iitjeepreparation #iitjeemaths #exponentialequations #exponents #exponential #exponent #systemsofequations #systems
#functionalequations #functions #function #maths #counting #sequencesandseries
#algebra #numbertheory #geometry #calculus #counting #mathcontests #mathcompetitions
via @RU-vid @Apple @Desmos @NotabilityApp @googledocs @canva
PLAYLISTS 🎵 :
Number Theory Problems: • Number Theory Problems
Challenging Math Problems: • Challenging Math Problems
Trigonometry Problems: • Trigonometry Problems
Diophantine Equations and Systems: • Diophantine Equations ...
Calculus: • Calculus

Опубликовано:

28 сен 2024

Ссылка:

Скачать:

Готовим ссылку...

Добавить в:

Мой плейлист

Посмотреть позже

Комментарии : 11

@mega_mango День назад

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

@surenderkumar4560 20 часов назад

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.

@pieintegration4083 День назад

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.

@leonidfedyakov366 14 часов назад

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.

@dondunno521 17 часов назад

-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?

@MrGeorge1896 День назад

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...

@TejasDhuri-p8z День назад

Very nice problem

@SyberMath День назад

Thanks

@Blaqjaqshellaq День назад

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

@bkkboy-cm3eb 21 час назад

S=1+1/3²+1/5²+···+1/(2n-1)²+··· 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