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Can the value of the Riemann zeta function at $n=2$ be derived from the Wallis formula for $pi$?


about the riemann zeta function and the prime counting functionNew tools for complex analysis and application to the Riemann Zeta function?Dirichlet series and Riemann zeta functionAre there any Riemann zeta like functions that may have nontrivial zeros on the critical line but only involves integers up to $N$?Contradicting statements about the Riemann zeta function at positive odd integersExtending the Riemann zeta function using Euler's Theorem.factor of Riemann Zeta product formula must be zero?Connection between Faulhauber's formula and Riemann zeta functionIrrationality of the values of the prime zeta functionWill this function / symbolic integral converge to the Riemann zeta zero counting function?













1












$begingroup$


It is well known that the Riemann zeta function, defined for all positive integers $n>1$ by
$$
zeta(n) = sum_m=1^infty m^-n
$$

takes the value $displaystyle fracpi^26$ at $n=2$. On the other hand, a product by Wallis converges to $displaystyle fracpi2$:
$$
prod _n=1^infty left(frac 2n2n-1cdot frac 2n2n+1right)=frac 21cdot frac 23cdot frac 43cdot frac 45cdot frac 65cdot frac 67cdot frac 87cdot frac 89cdots =frac pi 2.
$$



There exist many proofs of $zeta(2)=displaystyle fracpi^26$, but I have found none that involves the Wallis product.




  1. Can $zeta(2)$ be derived from the Wallis product?

  2. If so, can the general formula for $zeta(2n)$ by derived in a similar manner?










share|cite|improve this question











$endgroup$











  • $begingroup$
    Do you know this paper ? google.com/…
    $endgroup$
    – Claude Leibovici
    Mar 11 at 10:55










  • $begingroup$
    @ClaudeLeibovici I have given that a superficial reading, yes, but I haven't found such a theorem in the list it proposes.
    $endgroup$
    – Klangen
    Mar 11 at 11:57















1












$begingroup$


It is well known that the Riemann zeta function, defined for all positive integers $n>1$ by
$$
zeta(n) = sum_m=1^infty m^-n
$$

takes the value $displaystyle fracpi^26$ at $n=2$. On the other hand, a product by Wallis converges to $displaystyle fracpi2$:
$$
prod _n=1^infty left(frac 2n2n-1cdot frac 2n2n+1right)=frac 21cdot frac 23cdot frac 43cdot frac 45cdot frac 65cdot frac 67cdot frac 87cdot frac 89cdots =frac pi 2.
$$



There exist many proofs of $zeta(2)=displaystyle fracpi^26$, but I have found none that involves the Wallis product.




  1. Can $zeta(2)$ be derived from the Wallis product?

  2. If so, can the general formula for $zeta(2n)$ by derived in a similar manner?










share|cite|improve this question











$endgroup$











  • $begingroup$
    Do you know this paper ? google.com/…
    $endgroup$
    – Claude Leibovici
    Mar 11 at 10:55










  • $begingroup$
    @ClaudeLeibovici I have given that a superficial reading, yes, but I haven't found such a theorem in the list it proposes.
    $endgroup$
    – Klangen
    Mar 11 at 11:57













1












1








1


1



$begingroup$


It is well known that the Riemann zeta function, defined for all positive integers $n>1$ by
$$
zeta(n) = sum_m=1^infty m^-n
$$

takes the value $displaystyle fracpi^26$ at $n=2$. On the other hand, a product by Wallis converges to $displaystyle fracpi2$:
$$
prod _n=1^infty left(frac 2n2n-1cdot frac 2n2n+1right)=frac 21cdot frac 23cdot frac 43cdot frac 45cdot frac 65cdot frac 67cdot frac 87cdot frac 89cdots =frac pi 2.
$$



There exist many proofs of $zeta(2)=displaystyle fracpi^26$, but I have found none that involves the Wallis product.




  1. Can $zeta(2)$ be derived from the Wallis product?

  2. If so, can the general formula for $zeta(2n)$ by derived in a similar manner?










share|cite|improve this question











$endgroup$




It is well known that the Riemann zeta function, defined for all positive integers $n>1$ by
$$
zeta(n) = sum_m=1^infty m^-n
$$

takes the value $displaystyle fracpi^26$ at $n=2$. On the other hand, a product by Wallis converges to $displaystyle fracpi2$:
$$
prod _n=1^infty left(frac 2n2n-1cdot frac 2n2n+1right)=frac 21cdot frac 23cdot frac 43cdot frac 45cdot frac 65cdot frac 67cdot frac 87cdot frac 89cdots =frac pi 2.
$$



There exist many proofs of $zeta(2)=displaystyle fracpi^26$, but I have found none that involves the Wallis product.




  1. Can $zeta(2)$ be derived from the Wallis product?

  2. If so, can the general formula for $zeta(2n)$ by derived in a similar manner?







proof-writing riemann-zeta pi






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 11 at 11:58







Klangen

















asked Mar 11 at 10:10









KlangenKlangen

1,73711334




1,73711334











  • $begingroup$
    Do you know this paper ? google.com/…
    $endgroup$
    – Claude Leibovici
    Mar 11 at 10:55










  • $begingroup$
    @ClaudeLeibovici I have given that a superficial reading, yes, but I haven't found such a theorem in the list it proposes.
    $endgroup$
    – Klangen
    Mar 11 at 11:57
















  • $begingroup$
    Do you know this paper ? google.com/…
    $endgroup$
    – Claude Leibovici
    Mar 11 at 10:55










  • $begingroup$
    @ClaudeLeibovici I have given that a superficial reading, yes, but I haven't found such a theorem in the list it proposes.
    $endgroup$
    – Klangen
    Mar 11 at 11:57















$begingroup$
Do you know this paper ? google.com/…
$endgroup$
– Claude Leibovici
Mar 11 at 10:55




$begingroup$
Do you know this paper ? google.com/…
$endgroup$
– Claude Leibovici
Mar 11 at 10:55












$begingroup$
@ClaudeLeibovici I have given that a superficial reading, yes, but I haven't found such a theorem in the list it proposes.
$endgroup$
– Klangen
Mar 11 at 11:57




$begingroup$
@ClaudeLeibovici I have given that a superficial reading, yes, but I haven't found such a theorem in the list it proposes.
$endgroup$
– Klangen
Mar 11 at 11:57










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