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

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## Comments

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TopNewestCompletely 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 ?

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

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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.

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Also do Try my Integration Problems - here

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Wow! that's an amazing formula.

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

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

For example -

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

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

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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. $:)$

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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 !

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

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Nice Ronak.

Did you come up with using the integral.

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

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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.

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