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.
 A: 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
A: $\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.
