# A recursion formula for finding the value of Zeta function for all even values.

Initially I posted a proof for finding the value of $\zeta{(2)}$. Now I finally have generalized my approach and formed a recursion for finding the value of zeta function for all even values. Check this( it looks very complicated)

We have our recursion as :

$\displaystyle \sum _{ r=0 }^{ n }{ {}^{2n}{P}_{2r}\dfrac { { (-1) }^{ r }\zeta (2r+2) }{ { \pi }^{ 2r+2 } } } + { (-1) }^{ n }(2n)!\left(1-\dfrac { 1 }{ { 2 }^{ 2n+1 } } \right)\dfrac { \zeta (2n+2) }{ { \pi }^{ 2n+2 } } = \dfrac { 1 }{ 4(n+1) }$

This takes much simpler form if we take :

$\zeta (k)={ \pi }^{ k }{ a }_{ k }$

$\displaystyle \sum _{ r=0 }^{ n }{ {}^{2n}{P}_{2r} { (-1) }^{ r }{ a }_{ 2r+2 } } + { (-1) }^{ n }(2n)!\left(1-\frac { 1 }{ { 2 }^{ 2n+1 } } \right){ a }_{ 2n+2 } = \dfrac{1}{ 4(n+1) }$

You can try finding the values of zeta function from this.

where $\displaystyle {}^{n}{P}_{r} = \frac { n! }{ (n-r)! }$

Note by Ronak Agarwal
6 years, 3 months ago

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Completely Gone over my Head ! Since I didn't study zeta / gamma function . You guy's are awesome . Please tell me did you studied all such thing's from FIITJEE or from Internet ?

- 6 years, 3 months ago

These things are there on the internet as well in good calculus books for JEE.

- 6 years, 3 months ago

are u really 14 year teen ? Becoz when I was 14 I even don't know about JEE , And i didn;t here calculus word single time . It's quite surprising.

- 6 years, 3 months ago

Yes , I get that a lot .

Also do Try my Integration Problems - here

- 6 years, 3 months ago

Wow! that's an amazing formula.

I'll try to figure out the proof in meantime.

- 6 years, 3 months ago

Just a sidenote : for making large brackets use \left( and \right), that just makes the work look more cleaner.

For example -

$\left(1-\frac{1}{2^{2n+1}}\right)$

- 6 years, 3 months ago

I think you are a potential candidate for becoming a moderator , what do you think ?

- 6 years, 3 months ago

I would rather stay quiet in regard to this topic because I consider the Brilliant.org Team knows better than all of us that who are the better potential candidates to become a moderator.

- 6 years, 3 months ago

Haha nice answer , but if the Brilliant Team does ask the community for voting someone , my vote will definitely go to you ! You have all the qualities that a moderator should have and I personally think that after clearing JEE Advanced you should create your message board :)

- 6 years, 3 months ago

Oh, thanks! But I find that you're much more active than me on Brilliant and you take part in all kinds of discussions. Well, that's the quality, I guess, the Brilliant Team is looking for in their moderators, which basically means, you're more capable than me for becoming a moderator.

And for the JEE, my fingers are crossed and I hope for the best. I have an idea, how about starting a message board together (although separate but I mean at the same time) after we finish off with JEE.

And best of luck to you too. $:)$

- 6 years, 3 months ago

You see I'm not that active , I think I last came here on Brilliant at about 10 in the morning , right now I'm just answering the replies . Look at the picture below . I'll be back at around 10 pm or again next morning or so !!

Also I like your idea of starting a message board at the same time $\ddot\smile$ also I wouldn't mind the both of us sharing a message board haha !

- 6 years, 3 months ago

Nice formula Ronak :) I'll try to prove it some time later

- 6 years, 3 months ago

Nice Ronak.

Did you come up with using the integral.

- 6 years, 3 months ago

Honestly tell me the solution you posted to my problem was it your original approach.

- 6 years, 3 months ago

Actually, When I saw your approach of getting the Zeta function into the integral by substituting the exponential function into cos and using taylor series. I applied and it got me closer to the answer and I solved it.

- 6 years, 3 months ago