Talk:Dirichlet beta function

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Reciprocal of this function - is this worth adding to the main section?

It may be worth mentioniong that:

   1/beta(s) = sum(n=1..infinity((-1)^n * mu(n+1)/(2*n+1)^s));

which is valid for Re(s)>1. Hair Commodore (talk) 20:05, 28 July 2008 (UTC). Note that mu(*) is the Mobius mu function.[reply]

(It certainly converges to 4/Pi when s=1)

This formula is wrong! I tried checking it for 's'=1, 's'=2, and 's'=3, and it works at none of them! 81.102.15.200 (talk) 13:36, 18 August 2008 (UTC)[reply]

Apologies - I should have said that:

   1/beta(s) = sum(n=1..infinity,(((-1)^n * mu(2*n+1))/(2*n+1)^s))

in Maple notation - rather than what I typed previously.The anaonymous user who questioned it was quite correct. I made a wally error, so I have now corrected it. (It is still correct for 'Re'(s)>1) Hair Commodore (talk) 19:24, 18 August 2008 (UTC)[reply]

yes, from the Euler product I just added :

{\frac {1}{\beta (s)}}=\sum _{n=0}^{\infty }\mu _{2n+1}(2n+1)^{-s}(-1)^{n}

I'd still like to know if the Riemann hypothesis for

\beta (s)

would imply (or be implied by) the one for

\zeta (s)

, or if the two are completely independent. 78.227.78.135 (talk) 17:10, 4 January 2016 (UTC)[reply]