# This formula is on Wolfram Mathworld but I cannot find it anywhere else online?

$$\sum_{k=1}^{\infty}\ln\left[ \frac{(4k+1)^{1/(4k+1)^{n}}}{(4k-1)^{1/(4k-1)^{n}}} \right] = -\beta'(n)$$. Where $\beta$ is the Dirichlet Beta Function and $n$ is a positive integer.

I cannot find this cited anywhere nor values of the beta function derivative apart from at $-1,0,1$. How can I go about finding these things, I have searched googled and arxiv.

It is known that (uniformly and absolutely) $$\beta(n)=\sum^{\infty}_{k=1}\frac{\chi_4(k)}{k^n}\textrm{, }Re(n)>1.$$ Hence writing $1/k^n=e^{-n\log(k)}$, we have easily $$-\beta'(n)=\sum^{\infty}_{k=2}\frac{\chi_4(k)\log(k)}{k^n}.$$ But $$\chi_4(k)=\left\{ \begin{array}{cc} 0\textrm{ if }k\equiv 0 (mod)4\\ 1\textrm{ if }k\equiv 1 (mod)4\\ 0\textrm{ if }k\equiv 2 (mod)4\\ -1\textrm{ if }k\equiv 3 (mod)4 \end{array} \right\}.$$ Hence for $Re(n)>1$, we have $$-\beta'(n)=\sum^{\infty}_{k=1}\frac{\log(4k+1)}{(4k+1)^n}-\sum^{\infty}_{k=1}\frac{\log(4k-1)}{(4k-1)^n}=\sum^{\infty}_{k=1}\log\left(\frac{(4k+1)^{1/(4k+1)^{n}}}{(4k-1)^{1/(4k-1)^{n}}}\right).$$ QED

• I'd give the +50, but also asked about certain values and how they are computed Jul 1, 2018 at 14:03
• Here are some references: M. Waldschmidt, P. Moussa, J.-M. Luck, C. Itzykson (Eds.) "From Number Theory to Physics". Springer Verlag, Berlin, Heidelberg, New York, London, Paris, Tokyo, Hong Kong, Barcelona, Budapest. Printed (1992) in USA. Jul 1, 2018 at 15:08
• More: [2]: J.V. Armitage, W.F. Eberlein. 'Elliptic Functions'. Cambridge University Press. (2006)\\ [3]: Don. Zagier. 'Elliptic Modular Forms and Their Applications'. Available from [email protected] Jul 1, 2018 at 15:16
• [4]: J.M. Borwein, M.L. Glasser, R.C. McPhedran, J.G. Wan, I.J. Zucker. 'Lattice Sums Then and Now'. Cambridge University Press. New York, (2013). Jul 1, 2018 at 15:32

$\sum_{k=1}^{\infty}\ln\left[ \frac{(4k+1)^{1/(4k+1)^{n}}}{(4k-1)^{1/(4k-1)^{n}}} \right] = -\beta(n)$

I'll naively play with the left side and see what I can get.

The result is complicated expressions of dubious value, but here they are anyway.

$\begin{array}\\ t_k &=\ln\left[ \dfrac{(4k+1)^{1/(4k+1)^{n}}}{(4k-1)^{1/(4k-1)^{n}}} \right]\\ &=\ln((4k+1)^{1/(4k+1)^{n}})-\ln((4k-1)^{1/(4k-1)^{n}})\\ &=\dfrac1{(4k+1)^{n}}\ln(4k+1)-\dfrac1{(4k-1)^{n}}\ln(4k-1)\\ &=\dfrac{\ln(4k)}{(4k+1)^{n}}\ln(1+1/4k)-\dfrac{\ln(4k)}{(4k-1)^{n}}\ln(1-1/4k)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}(1+1/(4k))^{-n}\ln(1+1/(4k))-\dfrac{\ln(4k)}{(4k)^{n}}(1-1/(4k))^{-n}\ln(1-1/4k)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}\left((1+1/(4k))^{-n}\ln(1+1/(4k))-(1-1/(4k))^{-n}\ln(1-1/4k)\right)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}\left(f(-1/(4k))-f(1/(4k))\right)\\ \text{where}\\ f(x) &=(1-x)^{-n}\ln(1-x)\\ &=-\sum_{i=0}^{\infty} \binom{n+i-1}{k}x^i\sum_{j=1}^{\infty}\dfrac{x^j}{j}\\ &=-x\sum_{i=0}^{\infty} \binom{n+i-1}{k}x^i\sum_{j=0}^{\infty}\dfrac{x^j}{j+1}\\ &=-x\sum_{m=0}^{\infty} x^m\sum_{i=0}^{m}\binom{n+i-1}{i}\dfrac{1}{m-i+1}\\ &=-x\sum_{m=0}^{\infty} x^m\sum_{i=0}^{m}g(n, m, i)\\ \text{so}\\ t_k &=\dfrac{\ln(4k)}{(4k)^{n}}\left(f(-1/(4k))-f(1/(4k))\right)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}\left(-\dfrac{-1}{4k}\sum_{m=0}^{\infty} (-1)^m(4k)^{-m}\sum_{i=0}^{m}g(n, m, i)-\dfrac{1}{4k}\sum_{m=0}^{\infty} (4k)^{-m}\sum_{i=0}^{m}g(n, m, i)\right)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}\left(\dfrac{1}{4k}\sum_{m=0}^{\infty} (-1)^m(4k)^{-m}\sum_{i=0}^{m}g(n, m, i)-\dfrac{1}{4k}\sum_{m=0}^{\infty} (4k)^{-m}\sum_{i=0}^{m}g(n, m, i)\right)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}\left(\sum_{m=0}^{\infty}\dfrac{(4k)^{-m}}{4k}( (-1)^m-1)\sum_{i=0}^{m}g(n, m, i)\right)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}\left(\sum_{m=0}^{\infty}(4k)^{-m-1}( (-1)^m-1)\sum_{i=0}^{m}g(n, m, i)\right)\\ &=-2\dfrac{\ln(4k)}{(4k)^{n}}\left(\sum_{m=0}^{\infty}(4k)^{-(2m+1)-1}\sum_{i=0}^{2m+1}g(n, 2m+1, i)\right)\\ &=-2\dfrac{\ln(4k)}{(4k)^{n}}\left(\sum_{m=0}^{\infty}(4k)^{-2m-2}\sum_{i=0}^{2m+1}g(n, 2m+1, i)\right)\\ \end{array}$

At this point, I don't know what to do. However, I don't want to waste all this algebra, so I'll sumit this and hope that it might help someone else.